Quarry School

Local maxima and minima

Explain it like I am five

Think of one hill on a roller coaster. Its top can beat nearby heights even when a taller hill stands farther away. That is what local means: compare close neighbors on both sides. At a hilltop point (−2, 16), the local maximum is the height 16, and it occurs at input x = −2. A valley gives a local minimum in the same way. Relative maximum and relative minimum are other names for these ideas. Together they are local extrema. Rising then falling gives a hilltop; falling then rising gives a valley. The formal rule below also allows tied nearby heights, so a flat stretch needs separate attention.

−4−3−2−11234−40−32−24−16−8816243240outer samplelocal max (−2, 16)local min (2, −16)outer sample
The dots sample the given smooth curve f(x) = x3 − 12x; its stated turns are local, and the samples at ±4.5 visibly pass their heights.
Reminder
  • Ordered pairs. (−2, 16) means input −2, output 16. f(−2) = 16 says the same thing.
  • Comparing negatives. −1.633 is greater than −1.667 because it is closer to zero on the number line.
  • Squaring and substitution. (−2)2 = 4, but −(−2)2 = −4. Parentheses keep the substituted input together.
Why it works. Near a hilltop, points on the left are lower because the curve climbs toward the top; points on the right are lower because it falls away. This makes the top higher than nearby points on both sides. A valley reverses those comparisons. We compare within an open interval around the input, so an endpoint without domain points on both sides does not qualify under this text's definition. The value is an output because maximum and minimum describe height. The rule uses 'at least' and 'at most,' so equal nearby heights are allowed; strict hilltops and valleys are the turning cases.
RuleA local maximum at input b has f(b) ≥ f(x) for every x in some open interval around b contained in the domain. A local minimum has f(b) ≤ f(x) there. Its value is f(b); its location is b; its point is (b, f(b)). Equal nearby heights are allowed by these nonstrict inequalities. A strict local extremum uses > or < for every nearby x ≠ b.
The same idea, five ways
Say it

Say: 'local maximum of 16 at x = −2' and 'local minimum of −16 at x = 2.'

Write it

Write: 'The local maximum value is 16; it occurs at input −2, at point (−2, 16).'

In math
  • Location: b.
  • Value: f(b).
  • Point: (b, f(b)).
  • Local maximum: f(b) ≥ f(x) for all x in a small open domain interval around b.
  • Local minimum: f(b) ≤ f(x) in that neighborhood.
Like

The tallest or shortest house within your neighborhood, with nearby houses on both sides.

See it
24−11234left neighborvalue 3 at input 2right neighbor
The input 2 locates the vertex; the output 3 is its local maximum value.
The same idea, other ways
As a picture

For the given curve f(x) = x3 − 12x, the hilltop point is (−2, 16) and the valley point is (2, −16). The ends continue past those heights. These turns can be local extrema without being the whole graph's highest or lowest outputs.

−4−224−40−32−24−16−8816243240local max 16local min −16
These dots are exact samples of the given smooth curve, whose stated turns are at −2 and 2; the outer samples exceed the local heights.
As a neighborhood story

The tallest house on your block need not be the tallest in the whole city. A local maximum compares within one neighborhood of inputs. The neighborhood must include domain points to the left and right, as visiting houses on both sides of your address would.

13
An open neighborhood around b = 2 includes inputs on both sides of 2.
Why the comparison holds

For f(x) = 3 − (x − 2)2, the square is never negative. Thus f(x) ≤ 3 for every input, and f(2) = 3. At any other input the square is positive, so the height is lower than 3. This proves the hilltop comparison, rather than relying on two sample neighbors.

24−11234neighborb = 2neighbor
The parabola's vertex is (2, 3), and the square argument proves that other heights are lower.
With recorded costs

Among adjacent recorded years, the 2008 gas price $3.30 exceeds the 2007 and 2009 prices, while the 2009 price $2.41 is below the 2008 and 2010 prices. These are a sampled peak and dip useful for discussing observed costs. They are not a proof of continuous local extrema between recorded years.

input yearoutput price ($ per gallon)20072.8420083.320092.4120102.84↓ evaluate: input given, read the output below it
Adjacent sampled columns support an observed peak in 2008 and dip in 2009; unrecorded prices are unknown.
.1Local maximum: hilltop value, input, and point

A hilltop has an address and a height. The input is its address along the axis; the output is its height. A parabola is the U shaped graph of a quadratic, possibly opening down. Its vertex is its turning point.

  • Increasing then decreasing gives a strict local maximum.
  • A maximum value is an output; its input says where it occurs.
  • The formal nonstrict comparison also allows nearby equal outputs.
24−11234leftvertexright
Over the open interval (1, 3) around input 2, no point is higher than the vertex.
Worked exampleRead a hilltop

For f(x) = 3 − (x − 2)2, find the local maximum value, where it occurs, and its point.

24−11234232
The vertex's height is the maximum value, while its horizontal position is the location.
  1. f(2) = 3 − 02 = 3.Input 2 makes the square zero.
  2. At every different input, (x − 2)2 > 0, so f(x) < 3.A nonzero real number has a positive square.
  3. The value is 3, the location is x = 2, and the point is (2, 3).The ordered pair puts the input first and the output second.
Answer
  • Local maximum value: 3.
  • Location: x = 2.
  • Point: (2, 3).
Check f(1) = f(3) = 2 agrees with the pictured neighboring heights. The square argument supplies the proof for all nearby inputs.
.2Local minimum: valley value, input, and point

At a valley floor, both nearby sides lead upward. You still give its height and its address separately. Minimum describes the lower output, rather than the smallest input.

  • Decreasing then increasing gives a strict local minimum.
  • The value is the output at the valley.
  • The formal nonstrict comparison allows ties.
−2−22leftvertexright
The valley height is −2 at input −1.
Worked exampleRead a negative valley

Find the local minimum of f(x) = (x + 1)2 − 2.

−2−22−1−2−1
Negative −2 is lower than negative −1.
  1. f(−1) = 02 − 2 = −2.x + 1 equals zero at input −1.
  2. Every other input gives (x + 1)2 > 0, so its output exceeds −2.Adding a positive number to −2 raises it.
  3. Report local minimum of −2 at x = −1.The output supplies the value and the input supplies the location.
Answer
  • Local minimum of −2 at x = −1.
  • Point: (−1, −2).
Check f(−2) = −1 and f(0) = −1 are both above −2. The square comparison proves the result beyond these samples.
.3Ties on flat stretches and shoulders

Imagine a level shelf instead of a pointed hill. Its interior has nearby heights equal on both sides. 'At least as high' and 'at most as high' both permit equality, so the formal nonstrict definition can call those points both local types. They are not isolated turning points.

  • For a constant function, every interior input is a nonstrict local maximum and a nonstrict local minimum.
  • A flat stretch has no strict local extrema, because nearby equal heights fail the greater-than and less-than comparisons.
  • A shoulder at the edge of a flat stretch may satisfy one nonstrict comparison without an up-then-down turn. Compare both sides.
−221234samesamesame
Equal nearby heights satisfy the nonstrict rule but create no strict turn.
Worked exampleProving a strict local maximum on a downward parabola

The graph shows g(x) = −2(x − 1)2 + 6, defined for all real numbers. Points read from the graph are (−1, −2), (0, 4), (1, 6), (2, 4) and (3, −2). The curve keeps falling to the left of (−1, −2) and to the right of (3, −2). Show that g has a strict local maximum at x = 1. Give its value, its location and its point. Then explain why g has no other local maximum or minimum.

−224−4−22468(−1, −2)(0, 4)(1, 6)(2, 4)(3, −2)
Downward parabola g(x) = −2(x − 1)2 + 6 with the points (−1, −2), (0, 4), (1, 6), (2, 4) and (3, −2) marked. The highest point is the vertex (1, 6).
  1. Scan the listed points from left to right. The heights go −2, 4, 6, then fall to 4 and −2.A local maximum shows up where the graph changes from rising to falling. Here that change happens at x = 1.
  2. Check for flat stretches. The heights on either side of x = 1 are 4 and 4. Both are less than 6. The curve is a parabola, so it has no flat pieces.The nonstrict rule allows equal nearby heights. A strict maximum needs every nearby height to be lower, so we must rule out ties.
  3. Choose the open interval (0, 2) around b = 1. It lies inside the domain, which is all real numbers, so there are inputs on both sides of 1.A local extremum needs a neighbourhood inside the domain on both sides of b. An endpoint would not qualify.
  4. Prove the strict inequality algebraically. For any x ≠ 1, (x − 1)2 > 0. Multiplying by −2 gives −2(x − 1)2 < 0. Adding 6 gives g(x) < 6 = g(1).A strict local maximum needs g(1) > g(x) for every nearby x ≠ 1. The algebra shows this holds for every x ≠ 1 in (0, 2), and in fact for every real x ≠ 1.
  5. Read the location as the input x = 1 and the value as the output g(1) = −2(0)2 + 6 = 6. The point is (1, 6).The location is the horizontal input b, the value is f(b), and the point is the pair (b, f(b)).
  6. Look at every other input c ≠ 1. If c < 1, the graph is still rising at c, so slightly to the right of c it is higher. If c > 1, the graph is falling at c, so slightly to the left of c it is higher. No turn from falling to rising occurs anywhere.At every c ≠ 1 there is a higher point arbitrarily close, so c cannot be a local maximum. With no change from falling to rising, there is no local minimum.
Answer
g has a strict local maximum at x = 1 with value 6. The point is (1, 6). For every x ≠ 1, g(x) = 6 − 2(x − 1)2 < 6. g has no other local maximum and no local minimum.
Check Test inputs close to 1 on each side. g(0.9) = −2(−0.1)2 + 6 = 5.98 < 6 and g(1.1) = −2(0.1)2 + 6 = 5.98 < 6, which confirms the strict inequality. The listed points also agree with the formula: g(0) = −2(1) + 6 = 4 and g(3) = −2(4) + 6 = −2.

Work to write

  1. Rising then falling at x = 1: heights 4, 6, 4
  2. Neighbourhood (0, 2) lies in the domain on both sides of 1
  3. For x ≠ 1: (x − 1)2 > 0, so −2(x − 1)2 < 0, so g(x) < 6 = g(1)
  4. Strict local maximum at x = 1, value 6, point (1, 6)
  5. No other local extrema: no other change of direction

g has a strict local maximum at x = 1 with value 6. The point is (1, 6). For every x ≠ 1, g(x) = 6 − 2(x − 1)2 < 6. g has no other local maximum and no local minimum.

.4Read supplied estimates without a graphing calculator

A supplied graph may place a turn between labeled grid lines. Read its input and height as estimates. Checking nearby formula values can catch a misread sign or height, but a few samples cannot prove where every turn occurs.

  • The supplied graph of f(x) = 2x + x3 has a local maximum near (−2.449, −1.633) and a local minimum near (2.449, 1.633), with coordinates rounded.
  • The local maximum may be lower than a different local minimum, because the comparisons are in separate neighborhoods.
  • Input 0 is excluded, so behavior intervals must split there.
−6−4−2246−4−224estimated local maxestimated local min
Rounded sample dots represent the given two smooth branches; there is no point at 0 and no connecting curve across that gap.
Worked exampleChecking both branches of estimated heights for local extrema

A function f is defined only on the interval [-3, 3]. Between consecutive integer inputs its graph rises or falls steadily, with no extra bumps. Estimated heights read from the graph are: f(-3) = 2, f(-2) = 5, f(-1) = 7, f(0) = 5, f(1) = 1, f(2) = 3, f(3) = 6. Find every local maximum and local minimum of f. For each one, give its value and its location in a full sentence and state its point (b, f(b)). Explain why the endpoints x = -3 and x = 3 are not counted.

input xoutput f(x) (estimated height)−32−25−1705112336↓ evaluate: input given, read the output below it
Estimated heights of f on [-3, 3]. The highlighted inputs x = -1 (height 7) and x = 1 (height 1) are where the graph turns.
  1. Scan the heights from left to right: 2 → 5 → 7 is rising, 7 → 5 → 1 is falling, and 1 → 3 → 6 is rising again.Local extrema can only occur where the graph changes from rising to falling or from falling to rising.
  2. Mark the turns. At x = -1 the graph changes from rising to falling. At x = 1 it changes from falling to rising.Rising then falling identifies an isolated local maximum. Falling then rising identifies an isolated local minimum.
  3. Check both branches at x = -1. On the left, f(-2) = 5 < 7. On the right, f(0) = 5 < 7. Since the graph climbs steadily up to x = -1 and descends steadily after it, every x near -1 with x ≠ -1 has f(x) < 7.A local maximum needs f(b) ≥ f(x) on an open interval around b, on both sides. Here the inequality is strict, so the maximum is strict.
  4. Check both branches at x = 1. On the left, f(0) = 5 > 1. On the right, f(2) = 3 > 1. So every x near 1 with x ≠ 1 has f(x) > 1.A local minimum needs f(b) ≤ f(x) on both sides of b. Again there are no equal heights nearby, so the minimum is strict.
  5. Look for flat stretches or shoulders. The table shows no repeated heights next to each other, and the graph rises or falls steadily between the table inputs.The nonstrict rule would also allow equal nearby heights, so level stretches must be inspected. There are none here.
  6. Examine the endpoints. At x = -3 (height 2) the graph only extends to the right. At x = 3 (height 6) it only extends to the left.The domain is [-3, 3], so no open interval around -3 or around 3 lies inside the domain. Endpoints are considered separately, for absolute extrema.
  7. Read the horizontal input as the location and the vertical output as the value. The maximum has location -1 and value 7. The minimum has location 1 and value 1.The location is b, the value is f(b), and the point is (b, f(b)).
Answer
f has a local maximum value of 7 at x = -1, at the point (-1, 7). It has a local minimum value of 1 at x = 1, at the point (1, 1). The endpoints x = -3 and x = 3 are not local extrema because there is no domain interval on both sides of them.
Check Compare each candidate with its two neighbors. 5 < 7 > 5 confirms a peak at x = -1, and 5 > 1 < 3 confirms a valley at x = 1. The endpoint height 6 at x = 3 is larger than 1 and smaller than 7, but no input exists to its right, so it cannot be tested on both branches.

Work to write

  1. Rising then falling at x = -1: f(-2) = 5 < f(-1) = 7 > f(0) = 5
  2. Falling then rising at x = 1: f(0) = 5 > f(1) = 1 < f(2) = 3
  3. Local maximum value 7 at x = -1; point (-1, 7)
  4. Local minimum value 1 at x = 1; point (1, 1)
  5. x = -3 and x = 3 are endpoints with no interval on both sides, so they are not local extrema

f has a local maximum value of 7 at x = -1, at the point (-1, 7). It has a local minimum value of 1 at x = 1, at the point (1, 1). The endpoints x = -3 and x = 3 are not local extrema because there is no domain interval on both sides of them.

.5Retained sample data for the cubic picture

Sample dots are like a few photographed positions along a trail. They record exact or rounded heights at specific inputs. The given smooth curve supplies its behavior between those photographed positions.

  • These tables retain every plotted sample from the original cubic picture and add outer samples past the local heights.
  • Noninteger table heights are rounded to two decimal places.
  • Finite samples support the stated curve; they do not prove every nearby comparison or every turn.
input xoutput x³ − 12x, rounded−4.25−25.77−4−16−3.75−7.73−3.5−0.88−3.254.67−39−2.7512.20−2.514.38−2.2515.61
These retained sample values trace the given cubic to its left local maximum; displayed noninteger heights are rounded to two places.
Worked exampleLocating the turns of a cubic from a table of sample heights

The cubic f(x) = x3 − 12x + 4 is defined for all real numbers. Its heights at the integer inputs from −4 to 4 are: f(−4) = −12, f(−3) = 13, f(−2) = 20, f(−1) = 15, f(0) = 4, f(1) = −7, f(2) = −12, f(3) = −5, f(4) = 20. Between consecutive integer inputs the graph rises or falls steadily, with no extra bumps. Find every local maximum and local minimum of f in this window. Give the value and the location of each, and the point (b, f(b)).

input xoutput f(x) = x³ − 12x + 4−4−12−313−220−115041−72−123−5420↓ evaluate: input given, read the output below it
Sample heights of f(x) = x3 − 12x + 4 at x = −4 to 4. The turns are at x = −2 (height 20) and x = 2 (height −12).
  1. List the change in height between neighbouring samples: −12 → 13 (rise), 13 → 20 (rise), 20 → 15 (fall), 15 → 4 (fall), 4 → −7 (fall), −7 → −12 (fall), −12 → −5 (rise), −5 → 20 (rise).A turn shows up as a switch from rising to falling or from falling to rising, so we scan the sign of each change.
  2. At x = −2 the graph rises into it (13 → 20) and falls out of it (20 → 15). Mark x = −2 as a local maximum.Rising then falling identifies an isolated local maximum. Because the graph changes steadily between integers, every x in the open interval (−3, −1) other than −2 has f(x) < 20.
  3. At x = 2 the graph falls into it (−7 → −12) and rises out of it (−12 → −5). Mark x = 2 as a local minimum.Falling then rising identifies an isolated local minimum. On the open interval (1, 3), every x other than 2 has f(x) > −12.
  4. Check for flat stretches or shoulders. No two neighbouring samples are equal, and every other interior sample lies on a steady rise or a steady fall. So no other input qualifies under the nonstrict rule.The formal rule also allows equal nearby heights. Equal neighbours would be the only way to get an extra local extremum, and there are none here.
  5. Do not count x = −4 or x = 4. f(4) = 20 equals the local maximum value, but it is not a turn.The domain is all real numbers, so −4 and 4 are only the edges of the table, not ends of the graph. The cubic keeps falling to the left of −4 and keeps rising to the right of 4, so neither is a turn. Equal heights at different places do not create a new extremum.
  6. Read each location from the input column and each value from the output column.The location is the horizontal input b, the value is the vertical output f(b), and the point is (b, f(b)).
Answer
f has a local maximum value of 20 at x = −2, at the point (−2, 20). It has a local minimum value of −12 at x = 2, at the point (2, −12). Both are strict local extrema.
Check Use the derivative of the cubic: f′(x) = 3x2 − 12 = 3(x − 2)(x + 2), which is zero only at x = −2 and x = 2. f′ is positive for x < −2, negative for −2 < x < 2, and positive for x > 2. That confirms a maximum at −2 and a minimum at 2. Recompute the heights: f(−2) = −8 + 24 + 4 = 20 and f(2) = 8 − 24 + 4 = −12.

Work to write

  1. Heights rise from 13 to 20, then fall to 15, around x = −2, so x = −2 is a local maximum.
  2. Heights fall from −7 to −12, then rise to −5, around x = 2, so x = 2 is a local minimum.
  3. x = ±4 are only table edges; the domain is all reals and the graph is still falling or rising there, so they are not extrema.
  4. The local maximum value is 20 at x = −2, point (−2, 20).
  5. The local minimum value is −12 at x = 2, point (2, −12).

f has a local maximum value of 20 at x = −2, at the point (−2, 20). It has a local minimum value of −12 at x = 2, at the point (2, −12). Both are strict local extrema.

Strategy: step by step
  1. Scan the given graph for changes from rising to falling or falling to rising.
  2. Rising then falling identifies an isolated local maximum; falling then rising identifies an isolated local minimum.
  3. Also inspect flat stretches and shoulders when applying the formal nonstrict rule. Compare both sides, including any equal heights.
  4. Require a domain neighborhood on both sides. An end of the graph is considered separately for absolute extrema.
  5. Read the horizontal input as the location and the vertical output as the value.
  6. Give value and location in a full sentence; give the ordered pair if the question asks for the point.
Strategy
Find local values and locations
1
Does the graph rise and then fall?
YesThe turn gives a strict local maximum.
NoCheck for a valley or flat comparison.
↓
2
Does it fall and then rise?
YesThe turn gives a strict local minimum.
NoA flat stretch or shoulder can still satisfy the nonstrict inequalities; compare both sides.
↓
3
Are nearby domain points available on both sides of this input?
YesApply the local comparison within an open neighborhood.
NoThis text does not call the domain endpoint local; check it for an absolute extremum.
↓
4
Are you using a finite table of samples rather than a given full curve?
YesCall the result a sampled peak or dip unless additional information establishes the full nearby behavior.
NoRead the full given curve's neighborhood.
  1. Check changes of direction and any level stretches.
  2. Compare close domain points on both sides.
  3. Name maximum or minimum using the correct output comparison.
  4. Read the value and input separately and report both.
Worked exampleReading a local maximum from a downward-opening parabola

The graph shows f(x) = -(x - 2)2 + 5, defined for all real numbers. Points read from the graph are (0, 1), (1, 4), (2, 5), (3, 4) and (4, 1). The curve keeps falling to the left of (0, 1) and to the right of (4, 1). Find every local maximum and local minimum of f. Give each value and location in a sentence, and give the point.

−2246−4−2246(0, 1)(1, 4)(2, 5)(3, 4)(4, 1)
Graph of f(x) = -(x - 2)2 + 5 through (0, 1), (1, 4), (2, 5), (3, 4) and (4, 1). The curve rises to the local maximum at (2, 5) and then falls.
  1. Scan the graph from left to right. Through (0, 1) and (1, 4) up to (2, 5) the heights increase: 1, 4, 5. After x = 2 they decrease: 5, 4, 1.A local extremum shows up where the graph changes from rising to falling or from falling to rising.
  2. The graph rises and then falls at x = 2, so x = 2 is an isolated local maximum. Nowhere does the graph fall and then rise, so there is no isolated local minimum.Rising then falling marks a local maximum. Falling then rising would mark a local minimum.
  3. Check for flat stretches or shoulders. There are none. The curve is never level over an interval, and the points on each side of x = 2 are lower: f(1) = 4 < 5 and f(3) = 4 < 5. In general f(x) = 5 - (x - 2)2 < 5 for every x ≠ 2.The nonstrict rule asks us to compare both sides, including equal heights. Here every nearby height is strictly smaller, so the maximum is strict.
  4. Confirm the domain. f is defined for all real numbers, so an open interval such as (1, 3) lies around x = 2 inside the domain. The graph has no endpoints.A local extremum needs domain values on both sides. Ends of a graph are considered separately, for absolute extrema.
  5. Read the coordinates of the turning point. The horizontal input is 2, which is the location. The vertical output is f(2) = 5, which is the value.The location of an extremum is its input b, its value is f(b), and its point is (b, f(b)).
  6. State the result: f has a local maximum value of 5 at x = 2, at the point (2, 5). f has no local minimum.The answer must give the value and the location in a full sentence, plus the ordered pair the question asks for.
Answer
f has a local maximum value of 5 at x = 2. The point is (2, 5). f has no local minimum.
Check f(2) = -(0)2 + 5 = 5. On the open interval (1, 3), f(x) = 5 - (x - 2)2 ≤ 5, with equality only at x = 2. For example, f(1.5) = 4.75 and f(2.5) = 4.75, and both are less than 5. Since f rises for x < 2 and falls for x > 2, it never falls and then rises, so there is no local minimum.

Work to write

  1. The graph rises up to x = 2 and falls after it.
  2. f(x) ≤ f(2) = 5 for every x in the open interval (1, 3).
  3. Local maximum value 5 at x = 2, point (2, 5).
  4. There is no local minimum.

f has a local maximum value of 5 at x = 2. The point is (2, 5). f has no local minimum.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: Read a hilltop

For f(x) = 3 − (x − 2)2, find the local maximum value, where it occurs, and its point.

24−11234232
The vertex's height is the maximum value, while its horizontal position is the location.
  1. f(2) = 3 − 02 = 3.Input 2 makes the square zero.
  2. At every different input, (x − 2)2 > 0, so f(x) < 3.A nonzero real number has a positive square.
  3. The value is 3, the location is x = 2, and the point is (2, 3).The ordered pair puts the input first and the output second.
Answer
  • Local maximum value: 3.
  • Location: x = 2.
  • Point: (2, 3).
Check f(1) = f(3) = 2 agrees with the pictured neighboring heights. The square argument supplies the proof for all nearby inputs.
Rung 2Rung 2: Read a negative valley

Find the local minimum of f(x) = (x + 1)2 − 2.

−2−22−1−2−1
Negative −2 is lower than negative −1.
  1. f(−1) = 02 − 2 = −2.x + 1 equals zero at input −1.
  2. Every other input gives (x + 1)2 > 0, so its output exceeds −2.Adding a positive number to −2 raises it.
  3. Report local minimum of −2 at x = −1.The output supplies the value and the input supplies the location.
Answer
  • Local minimum of −2 at x = −1.
  • Point: (−1, −2).
Check f(−2) = −1 and f(0) = −1 are both above −2. The square comparison proves the result beyond these samples.
Rung 3Reading a local maximum from a downward parabola with its vertex left of the y-axis

The graph shows h(x) = −(x + 1)2 + 3, defined for all real numbers. Points read from the graph are (−3, −1), (−2, 2), (−1, 3), (0, 2) and (1, −1). The curve keeps falling to the left of (−3, −1) and to the right of (1, −1). Find every local maximum and every local minimum of h. For each local maximum, state its value and its location, and give the point.

−4−22−6−4−224(−3, −1)(−2, 2)(−1, 3)(0, 2)(1, −1)
Downward parabola h(x) = −(x + 1)2 + 3 with the marked points (−3, −1), (−2, 2), (−1, 3), (0, 2) and (1, −1). It peaks at (−1, 3).
  1. Scan the listed points from left to right: −1, 2, 3, 2, −1. The heights rise from x = −3 to x = −1 and fall from x = −1 to x = 1.A local extremum shows up where the graph changes direction, so the first job is to find where rising turns into falling or falling turns into rising.
  2. The only change of direction is at x = −1, where the graph goes from rising to falling. This marks a local maximum at x = −1.Rising then falling identifies an isolated local maximum.
  3. Look for flat stretches or shoulders. There are none: the heights on the left of x = −1 (for example h(−2) = 2) and on the right (for example h(0) = 2) are both below 3, and the curve keeps falling beyond the listed points.The nonstrict rule h(b) ≥ h(x) must be checked on both sides of b, and any equal nearby heights would also need attention.
  4. Confirm with the formula: h(x) = −(x + 1)2 + 3 ≤ 3 = h(−1) for every x, because (x + 1)2 ≥ 0. Equality holds only at x = −1, so the open interval (−2, 0) works, and the maximum is strict.The formal definition needs h(b) ≥ h(x) for all x in some open interval around b. It is strict when the inequality is > for every nearby x ≠ b.
  5. Check for a local minimum. The graph never changes from falling to rising, and the domain is all real numbers, so there are no endpoints to consider. h has no local minimum.A local minimum needs falling then rising, with domain on both sides of the point. There is no such place.
  6. Read the location from the horizontal axis, x = −1, and the value from the vertical axis, h(−1) = 3. The point is (−1, 3).The location is the input b, the value is the output h(b), and the point is (b, h(b)).
Answer
h has a local maximum value of 3, located at x = −1, so the point is (−1, 3). This local maximum is strict. h has no local minimum.
Check Substitute x = −1: h(−1) = −(0)2 + 3 = 3. Neighbors: h(−1.5) = −0.25 + 3 = 2.75 and h(−0.5) = −0.25 + 3 = 2.75. Both are less than 3, which agrees with a strict local maximum at x = −1. A downward parabola (leading coefficient −1) has exactly one turning point, its vertex (−1, 3), so no local minimum is missing.

Work to write

  1. The graph rises up to x = −1 and falls after x = −1.
  2. h(x) = −(x + 1)2 + 3 ≤ 3 = h(−1) for all x, so h(−1) ≥ h(x) on an open interval around −1.
  3. h has a local maximum value of 3 at x = −1.
  4. The point is (−1, 3).
  5. h has no local minimum.

h has a local maximum value of 3, located at x = −1, so the point is (−1, 3). This local maximum is strict. h has no local minimum.

Rung 4Proving a local minimum with the inequality on a restricted domain

A function is given by f(x) = (x − 1)2 − 4, defined only on the interval [−2, 3]. Points read from its graph are (−2, 5), (−1, 0), (0, −3), (1, −4), (2, −3) and (3, 0). The graph falls from x = −2 to x = 1 and rises from x = 1 to x = 3. Find every local maximum and local minimum of f. Justify each one with the defining inequality and say whether it is strict. For each, give the value, the location and the point.

−224−4−2246(-2, 5)(-1, 0)(0, -3)(1, -4)(2, -3)(3, 0)
The parabola y = (x − 1)2 − 4 with the points (−2, 5), (−1, 0), (0, −3), (1, −4), (2, −3) and (3, 0) marked. The function's domain is only the part from x = −2 to x = 3, and its lowest point is (1, −4).
  1. Scan the listed heights 5, 0, −3, −4, −3, 0. The graph falls from (−2, 5) down to (1, −4) and then rises to (3, 0).Local extrema can only occur where the graph changes direction, or along flat stretches.
  2. The only change of direction is at x = 1, from falling to rising. Mark x = 1 as a candidate local minimum. The graph has no change from rising to falling, so there is no interior candidate for a local maximum.Falling then rising signals a local minimum. Rising then falling would signal a local maximum.
  3. Apply the inequality at b = 1. f(x) − f(1) = [(x − 1)2 − 4] − (−4) = (x − 1)2. This is ≥ 0, so f(1) ≤ f(x) for every x in the open interval (0, 2), which lies inside [−2, 3].A local minimum at b requires f(b) ≤ f(x) for all x in some open interval around b that is contained in the domain.
  4. Check strictness. For x ≠ 1, (x − 1)2 > 0, so f(1) < f(x) for every nearby x ≠ 1. There is no flat stretch or equal nearby height.A strict local minimum needs < for every nearby x other than b. A square of a nonzero number is positive.
  5. Examine the ends x = −2 and x = 3. No open interval around either end fits inside [−2, 3], because the domain stops on one side. So neither end is a local maximum or local minimum. The height f(−2) = 5 is handled separately, as the absolute maximum.The local definition requires domain points on both sides of b. Ends are considered only for absolute extrema.
  6. Read the horizontal input as the location: 1. Read the vertical output as the value: f(1) = (1 − 1)2 − 4 = −4.The location is b, the value is f(b), and the point is (b, f(b)).
Answer
f has a strict local minimum of value −4 at x = 1, at the point (1, −4). f has no local maximum. The endpoints x = −2 and x = 3 are not local extrema because no open interval around them lies in the domain.
Check Test inputs near 1 on both sides. f(0.9) = (−0.1)2 − 4 = −3.99 and f(1.1) = (0.1)2 − 4 = −3.99. Both are greater than −4, as the strict inequality predicts. The listed points (0, −3) and (2, −3) are also higher than (1, −4).

Work to write

  1. Graph falls on [−2, 1] and rises on [1, 3]; the only turn is at x = 1.
  2. f(x) − f(1) = (x − 1)2 ≥ 0, so f(1) ≤ f(x) on (0, 2) ⊂ [−2, 3].
  3. (x − 1)2 > 0 for x ≠ 1, so the local minimum is strict.
  4. x = −2 and x = 3 are endpoints: no open interval in the domain on both sides, so they are not local extrema.
  5. Local minimum value −4 at x = 1; point (1, −4). No local maximum.

f has a strict local minimum of value −4 at x = 1, at the point (1, −4). f has no local maximum. The endpoints x = −2 and x = 3 are not local extrema because no open interval around them lies in the domain.

Rung 5Checking a claimed peak against both neighbors

A graph of y = f(x) on the domain [-3, 3] is made of straight segments joining these points, read from the graph: (-3, 1), (-2, 4), (-1, 6), (0, 3), (1, -2), (2, 0), (3, 5). A student says f has a local maximum at x = -2 because f(-2) = 4 is higher than f(-3) = 1. Check this claim by comparing heights on both sides. Then find every local maximum and local minimum of f, giving the value and location of each in a sentence and the point as an ordered pair.

input xoutput f(x)−31−24−16031−22035↓ evaluate: input given, read the output below it
Heights read from the graph of f at x = -3 through 3. The peak f(-1) = 6 and the valley f(1) = -2 are marked.
  1. Check the student's claim at x = -2 on the left side. f(-3) = 1 < 4 = f(-2), and the segment from x = -3 to x = -2 rises.A local maximum needs f(-2) ≥ f(x) for all nearby x on both sides. The left side passes.
  2. Check the right side of x = -2. f(-1) = 6 > 4. Every x just to the right of -2 lies on the rising segment toward (-1, 6), so f(x) > 4 there.One branch higher than f(-2) in every open interval around -2 breaks the condition f(-2) ≥ f(x). The claim fails, and x = -2 is not a local maximum.
  3. Scan the heights in order: 1, 4, 6, 3, -2, 0, 5. The graph rises to x = -1, falls to x = 1, then rises to x = 3.Local extrema in the interior sit where rising changes to falling or falling changes to rising.
  4. At x = -1, the left neighbor is f(-2) = 4 < 6 and the right neighbor is f(0) = 3 < 6. Both segments slope down away from (-1, 6), so f(x) < 6 for nearby x ≠ -1.Rising then falling, checked on both branches, gives a strict local maximum.
  5. At x = 1, the left neighbor is f(0) = 3 > -2 and the right neighbor is f(2) = 0 > -2. Both segments slope up away from (1, -2), so f(x) > -2 for nearby x ≠ 1.Falling then rising, checked on both branches, gives a strict local minimum.
  6. Exclude x = -3 and x = 3.They are ends of the domain, so no open interval around them lies inside the domain. Ends are considered separately, for absolute extrema.
  7. Read each location from the horizontal input and each value from the vertical output.The location is b, the value is f(b), and the point is (b, f(b)).
Answer
The student's claim is wrong: f(-1) = 6 > f(-2) = 4, so x = -2 is not a local maximum. f has a local maximum value of 6 at x = -1, at the point (-1, 6). It has a local minimum value of -2 at x = 1, at the point (1, -2). There are no other local extrema.
Check Each candidate is compared with its neighbor on both sides. For x = -1: 4 < 6 and 3 < 6. For x = 1: 3 > -2 and 0 > -2. At x = -2, the right neighbor 6 is greater than 4, so that point fails. The domain ends at x = -3 and x = 3 are left out because they have no neighbors on both sides.

Work to write

  1. Right of x = -2: f(-1) = 6 > 4, so x = -2 is not a local max
  2. f(-2) = 4 < 6 and f(0) = 3 < 6, so local max at x = -1
  3. Local maximum value 6 at x = -1, point (-1, 6)
  4. f(0) = 3 > -2 and f(2) = 0 > -2, so local min at x = 1
  5. Local minimum value -2 at x = 1, point (1, -2)
  6. x = -3 and x = 3 are domain ends, not local extrema

The student's claim is wrong: f(-1) = 6 > f(-2) = 4, so x = -2 is not a local maximum. f has a local maximum value of 6 at x = -1, at the point (-1, 6). It has a local minimum value of -2 at x = 1, at the point (1, -2). There are no other local extrema.

Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The local maximum is x = 1.
1 is the input location. In the worked graph the local maximum output is 2.
✓ Instead: Write 'local maximum of 2 at x = 1'; its point is (1, 2).
✗ Not this: Every local maximum must exceed every local minimum.
The supplied rational graph has local maximum about −1.633 on one branch and local minimum about 1.633 on another.
✓ Instead: Local comparisons use one neighborhood, so extrema in different neighborhoods need not be ordered that way.
✗ Not this: Every formal local extremum must be an isolated turn.
The nonstrict rule allows equality. Every interior input of a constant function satisfies both local comparisons.
✓ Instead: Distinguish nonstrict local extrema from strict hilltops, valleys, and isolated turning points.
✗ Not this: Two lower sampled neighbors prove a local maximum.
Unmeasured inputs between those samples could have higher outputs.
✓ Instead: Use a given full curve or an argument applying to every nearby input; use finite samples as supporting checks.
Tips and tricks
  • For a value, give the output and where it occurs. For a point, give (input, output).
  • In this text, a domain endpoint without nearby domain points on both sides is not local. It can still be an absolute extremum.
  • Graph exercises commonly emphasize isolated turns. If a flat stretch appears, state how the nonstrict rule treats it.
  • Only the TI-30XIIS is allowed on this course's exam. Read the supplied graph; nearby substitutions support a suspected shape but cannot certify all turns.
Trap. Saying 'the maximum is x = 1'. The maximum is the output, 2; x = 1 is where it happens. On the exam write both: 'local maximum of 2 at x = 1'.