When an endpoint is a letter, the answer is an expression

This asks for the average rate from a fixed input to any different input b. Find it for f(x) = 4x + 9, starting at −2.

1. f(−2) = 4(−2) + 9 = −8 + 9 = 1; f(b) = 4b + 9.The first output is numerical; the second keeps the endpoint letter.
2. Output change: f(b) − f(−2) = (4b + 9) − 1 = 4b + 8.Subtract the entire starting output.
3. Input change: b − (−2) = b + 2.The input change follows the same endpoint order.
4. 4b + 8 = 4(b + 2), so 4(b + 2)b + 2$\frac{4(b+2)}{b+2}$ = 4, with b ≠ −2.The factor b + 2 multiplies the whole top. The original bottom excludes b = −2.

Let f(x) = x²$x^2$ − 3x − 10. (a) Write the average rate of change of f from x = 5 to x = a, where a ≠ 5, as a simplified expression in a. (b) The average rate of change on [5, a] is 9. Find a.

1. f(5) = 5²$5^2$ − 3(5) − 10 = 25 − 15 − 10 = 0.The fixed numerical endpoint is evaluated first. Its output is 0, so subtracting it will not change the numerator.
2. f(a) = a²$a^2$ − 3a − 10.Replacing x with the endpoint letter a gives the other output.
3. Rate = f(a) − f(5)a − 5$\frac{f\left(a\right)-f\left(5\right)}{a-5}$ = (a² − 3a − 10) − 0a − 5$\frac{(a^2-3a-10)-0}{a-5}$ = a² − 3a − 10a − 5$\frac{a^2-3a-10}{a-5}$, with a ≠ 5.Average rate of change is the change in output divided by the matching change in input, taken in the same order. The denominator cannot be 0.
4. Factor the numerator: a²$a^2$ − 3a − 10 = (a − 5)(a + 2). Check by multiplying back: (a − 5)(a + 2) = a²$a^2$ + 2a − 5a − 10 = a²$a^2$ − 3a − 10.We need two numbers with product −10 and sum −3. Those are −5 and 2. Multiplying back confirms the factoring.
5. Rate = (a − 5)(a + 2)a − 5$\frac{(a-5)(a+2)}{a-5}$ = a + 2, for a ≠ 5.(a − 5) multiplies the whole numerator and matches the denominator, so it cancels. The restriction a ≠ 5 from the original quotient still applies.
6. Set a + 2 = 9, so a = 7. Since 7 > 5, the interval is [5, 7].The given rate equals the simplified expression. The solution a = 7 is not 5, so it is allowed, and it gives the sorted interval [5, a].
7. Check in the original quotient: f(7) = 49 − 21 − 10 = 18, so 18 − 07 − 5$\frac{18-0}{7-5}$ = 182$\frac{18}{2}$ = 9.The endpoint must be checked in the original, unsimplified quotient to confirm the rate is 9.

This asks for the rate expression in b, using a fixed starting input 6. Find the average rate of f(x) = x²$x^2$ + 5 from 6 to b.

1. f(6) = 6²$6^2$ + 5 = 36 + 5 = 41; f(b) = b²$b^2$ + 5.Evaluate the fixed endpoint and substitute the letter at the other endpoint.
2. f(b) − f(6) = (b²$b^2$ + 5) − 41 = b²$b^2$ − 36.Subtract the whole starting output; the constants combine: 5 − 41 = −36.
3. b²$b^2$ − 36 = b²$b^2$ − 6²$6^2$ = (b − 6)(b + 6).Multiplying these two factors gives b²$b^2$ + 6b − 6b − 36; the middle terms cancel.
4. (b − 6)(b + 6)b − 6$\frac{(b-6)(b+6)}{b-6}$ = b + 6, with b ≠ 6.The input change is b − 6, and that nonzero factor multiplies the whole numerator.

This asks for a rate expression rather than one rate number. Find the average rate of f(x) = 3x²$x^2$ − 5 on [2, b], with b > 2.

1. f(2) = 3(2)²${)}^2$ − 5 = 3(4) − 5 = 7; f(b) = 3b²$b^2$ − 5.Evaluate the fixed input 2 and then substitute the endpoint b.
2. f(b) − f(2) = (3b²$b^2$ − 5) − 7 = 3b²$b^2$ − 12.Subtract the whole starting output; −5 − 7 = −12.
3. 3b²$b^2$ − 12 = 3(b²$b^2$ − 4).Both terms contain a factor of 3.
4. b²$b^2$ − 4 = (b − 2)(b + 2).Multiplying back gives b²$b^2$ + 2b − 2b − 4 = b²$b^2$ − 4.
5. 3(b − 2)(b + 2)b − 2$\frac{3(b-2)(b+2)}{b-2}$ = 3(b + 2) = 3b + 6, with b ≠ 2.The input difference b − 2 is a nonzero factor of the entire top.

Let f(x) = (x − 1)²${)}^2$ − 3, which expands to f(x) = x²$x^2$ − 2x − 2. (a) Write the average rate of change of f between x = 5 and x = a as a simplified expression in a, with a ≠ 5. (b) The average rate of change on [5, a] is 10. Find a.

1. f(5) = 5²$5^2$ − 2(5) − 2 = 25 − 10 − 2 = 13.The fixed numerical endpoint is evaluated first, so one output is a plain number.
2. f(a) = a²$a^2$ − 2a − 2.Substituting the letter a for x gives the other output as an expression.
3. Rate = f(a) − f(5)a − 5$\frac{f\left(a\right)-f\left(5\right)}{a-5}$ = (a² − 2a − 2) − 13a − 5$\frac{(a^2-2a-2)-13}{a-5}$ = a² − 2a − 15a − 5$\frac{a^2-2a-15}{a-5}$, with a ≠ 5.The average rate is the change in output over the matching change in input, taken in the same order. a ≠ 5 keeps the denominator nonzero.
4. Factor a²$a^2$ − 2a − 15. Look for two numbers with product −15 and sum −2. These are −5 and 3, so a²$a^2$ − 2a − 15 = (a − 5)(a + 3). Check: (a − 5)(a + 3) = a²$a^2$ + 3a − 5a − 15 = a²$a^2$ − 2a − 15.This is the sum-and-product puzzle for a monic quadratic. Multiplying back confirms the factorization is correct.
5. (a − 5)(a + 3)a − 5$\frac{(a-5)(a+3)}{a-5}$ = a + 3, for a ≠ 5.(a − 5) multiplies the whole numerator and matches the denominator, so it cancels. The restriction a ≠ 5 from the original quotient still applies.
6. Set a + 3 = 10, so a = 7. Since 7 > 5, the interval is [5, 7].The given rate equals the simplified expression. The value a = 7 is allowed because it is not 5.

Let f(x) = x²$x^2$ − 6x + 4, which can also be written f(x) = (x − 3)²${)}^2$ − 5. (a) Write the average rate of change of f between the inputs 5 and a, where a ≠ 5, as a simplified expression in a. (b) The average rate of change of f between 5 and a equals 3. Find a and state the interval, written with the smaller endpoint first.

1. f(5) = 5²$5^2$ − 6(5) + 4 = 25 − 30 + 4 = −1The fixed endpoint is a number, so its output can be computed directly.
2. f(a) = a²$a^2$ − 6a + 4Substituting the letter a for x gives the output at the other endpoint.
3. rate = f(a) − f(5)a − 5$\frac{f\left(a\right)-f\left(5\right)}{a-5}$ = (a² − 6a + 4) − (−1)a − 5$\frac{(a^2-6a+4)-(-1)}{a-5}$ = a² − 6a + 5a − 5$\frac{a^2-6a+5}{a-5}$Average rate of change is the change in output divided by the matching change in input, with the inputs in the same order as the outputs.
4. a²$a^2$ − 6a + 5 = (a − 1)(a − 5). Check: (a − 1)(a − 5) = a²$a^2$ − 5a − a + 5 = a²$a^2$ − 6a + 5Factoring shows whether the numerator contains the factor (a − 5). Multiplying back confirms the factoring.
5. rate = (a − 1)(a − 5)a − 5$\frac{(a-1)(a-5)}{a-5}$ = a − 1, for a ≠ 5(a − 5) multiplies the whole numerator, so it cancels. The restriction a ≠ 5 stays because the original quotient is undefined at a = 5.
6. a − 1 = 3, so a = 4Setting the simplified rate equal to the given value gives a linear equation in a.
7. a = 4 satisfies a ≠ 5. Since 4 < 5, the sorted interval is [4, 5].The solution must satisfy the original restriction. An interval is written with the smaller endpoint first.

This asks for an endpoint giving a specified average rate. Let g(x) = 12x$\frac{12}{x}$. Starting at 3, find b > 3 so the average rate on [3, b] is −1.

1. Require b ≠ 0 for g(b), and b ≠ 3 for the input change. g(3) = 123$\frac{12}{3}$ = 4; g(b) = 12b$\frac{12}{b}$.Both the function output and the rate quotient need nonzero bottoms.
2. g(b) − g(3) = 12b$\frac{12}{b}$ − 4 = 12b$\frac{12}{b}$ − 4bb$\frac{4b}{b}$ = 12 − 4bb$\frac{12-4b}{b}$.Write the whole number 4 over bottom b by multiplying it by bb$\frac{b}{b}$ = 1.
3. 12 − 4b = −4(b − 3).Multiplying −4 through the parentheses gives −4b + 12, the same numerator.
4. The rate is −4(b − 3)b$\frac{-4(b-3)}{b}$ ÷ (b − 3) = −4(b − 3)b$\frac{-4(b-3)}{b}$ · 1b − 3$\frac{1}{b-3}$ = −4b$\frac{-4}{b}$.Dividing by b − 3 means multiplying by its reciprocal; the factor cancels only because b ≠ 3.
5. Set −4b$\frac{-4}{b}$ = −1, then multiply both sides by b: −4 = −b.This finds the input that gives rate −1; multiplying by b is allowed because b ≠ 0.
6. Multiply both sides by −1: b = 4.The same multiplication preserves the equality and makes b positive.
7. 4 > 3, 4 ≠ 0, and 4 ≠ 3.The candidate satisfies the requested interval and both original restrictions.