Graph each piece only where its condition allows it

Graph F(x) = x for x < 0 and F(x) = 2x for x ≥ 0. State domain and range.

1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
2. Draw the line y = x only left of 0 with an open point at (0, 0).The first branch excludes its boundary and gives negative outputs for negative inputs.
3. Draw the line y = 2x on and right of 0 with a closed point at (0, 0), passing through (1, 2).The second branch includes the boundary and has twice the input as its output.
4. Keep the included point at (0, 0) where the branches meet.An open marker from the first piece does not remove a point included by the second piece.
5. Take the domain union (−∞, 0) ∪ [0, ∞).All negative and nonnegative inputs are covered.
6. Take the range union (−∞, 0) ∪ [0, ∞).The first piece reaches every negative output; the second reaches every nonnegative output.

Graph G(x) = x − 2 for x < 4 and G(x) = x + 3 for x ≥ 4. Find domain and range.

1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
2. Draw y = x − 2 left of 4, ending at the open point (4, 2).Inputs below 4 use the first branch. At 4 it would give 4 − 2 = 2, but 4 is not below 4, so that point is hollow and these outputs stay strictly below 2.
3. Draw y = x + 3 on and right of 4, starting at the closed point (4, 7) and passing through (5, 8).The second branch includes 4, where 4 + 3 = 7, and it produces outputs at least 7.
4. The input intervals join to all real numbers.Every real input is either below 4 or at least 4.
5. The output intervals remain (−∞, 2) and [7, ∞).No included point reaches a height from 2 up to but not including 7.
6. Join the output intervals with ∪.A union preserves the missing heights rather than filling them.

Let f(x) = x²$x^2$ if x ≤ 1, f(x) = 3 if 1 < x ≤ 2, and f(x) = x if x > 2. Graph it, find f(1), f(1.5), f(2) and f(4), and state domain and range.

1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
2. Draw y = x²$x^2$ only for x ≤ 1, with a closed point at (1, 1).The first condition includes its boundary; the square curve has its lowest point at (0, 0).
3. Draw y = 3 on 1 < x ≤ 2, with an open point at (1, 3) and a closed point at (2, 3).This condition excludes its left endpoint and includes its right endpoint.
4. Draw y = x only for x > 2, beginning with an open point at (2, 2).The third condition excludes 2, so its line does not assign an output there.
5. At x = 1, choose the first branch and calculate 1²$1^2$ = 1.The ≤ 1 condition owns this boundary.
6. At x = 1.5, choose the middle branch and get 3.1 < 1.5 ≤ 2.
7. At x = 2, also choose the middle branch and get 3.The middle condition includes equality at 2; the final branch does not.
8. At x = 4, choose the last branch and get 4.4 satisfies x > 2.
9. Take the input union (−∞, 1] ∪ (1, 2] ∪ (2, ∞) = (−∞, ∞).Every real input belongs to one of these three intervals, with no gap.
10. The first branch already reaches every y ≥ 0, and neither other branch goes below zero.For any y ≥ 0 the allowed input −y$\sqrt{y}$ supplies that output through the first branch.
11. Write range [0, ∞).Union with {3} and (2, ∞) adds no new values beyond the first branch's range.

Let H(x) = x if x < −1 and H(x) = x + 2 if x ≥ 2. Graph it and state its domain and range.

1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
2. Draw y = x for x < −1, ending with an open point at (−1, −1).The first branch excludes its right boundary and extends downward to the left.
3. Leave all inputs from −1 through values below 2 blank.Neither condition assigns an output to those inputs.
4. Draw y = x + 2 for x ≥ 2, starting with a closed point at (2, 4) and passing through (3, 5).The second condition includes 2 and gives output 2 + 2 = 4 there.
5. Write domain (−∞, −1) ∪ [2, ∞).These are exactly the two allowed input intervals.
6. Write range (−∞, −1) ∪ [4, ∞).The first line reaches every height below −1; the second reaches every height at least 4.

Let Q(x) = 5 if x < −4 and Q(x) = x + 4$\sqrt{x+4}$ if x ≥ −4. Graph the two assigned pieces and find domain and range. You are drawing each shape only for its own input condition, then collecting the whole graph’s two shadows.

1. For x < −4, draw a horizontal ray at height 5, with an open point at (−4, 5) and continuation to the left.The constant rule gives one height on its assigned inputs, and the strict condition excludes the boundary.
2. For x ≥ −4, draw the root curve beginning at the closed point (−4, 0). At x = 0 it passes through (0, 2), and at x = 5 it passes through (5, 3).Substitution gives 0$\sqrt{0}$ = 0, 4$\sqrt{4}$ = 2, and 9$\sqrt{9}$ = 3. The second condition includes the starting input and its root is real for every assigned input.
3. Take the input union (−∞, −4) ∪ [−4, ∞) = (−∞, ∞).Every input below the boundary uses the constant, and every input at or above it uses the root, so none is missing.
4. The constant branch gives {5}. The root branch gives [0, ∞), because any target y ≥ 0 comes from x = y²$y^2$ − 4 ≥ −4.The nonnegative root outputs are all attained by the constructed input.
5. The full output union is {5} ∪ [0, ∞) = [0, ∞).Output 5 already lies in the root range, so including the constant branch adds no new height.

Let P(x) = x³$x^3$ if x < −3, P(x) = −7 if −3 < x < 9, and P(x) = x − 5$\sqrt{x-5}$ if x > 9. Graph the pieces and find domain and range. You are retaining the branch shapes only on the stated intervals, including any missing inputs or isolated output heights.

1. Draw the cube curve only for x < −3. At its right boundary the formula would give (−3)³${)}^3$ = −27, so place an open point at (−3, −27).The strict condition excludes −3. Every earlier input gives a cube below −27.
2. Draw a horizontal segment at height −7 only for −3 < x < 9, with both (−3, −7) and (9, −7) open.Both comparisons are strict, although every interior input supplies the actual output −7.
3. For x > 9, the root curve’s boundary location is (9, 2), because 9 − 5$\sqrt{9-5}$ = 4$\sqrt{4}$ = 2. Leave that point open, and draw the curve through (14, 3), continuing rightward.Only inputs above 9 belong to this branch; they make the root inside greater than 4, so their root outputs are strictly greater than 2.
4. Keep both missing input values and write (−∞, −3) ∪ (−3, 9) ∪ (9, ∞).No condition includes −3 or 9; each other real input belongs to exactly one branch.
5. The cube branch reaches exactly (−∞, −27): for any y < −27, the input ³y$\sqrt[3]{y}$ is below −3 and cubes to y. The middle branch reaches the single value −7.Cubing preserves input order, and the constant branch has only one output despite its many inputs.
6. The root branch reaches (2, ∞): for any target y > 2, choose x = y²$y^2$ + 5 > 9; then x − 5$\sqrt{x-5}$ = y.The target is positive, so its square exceeds 4 and the principal root returns y. This proves every claimed root height is attained.
7. Combine the three output sets as (−∞, −27) ∪ [−7, −7] ∪ (2, ∞).The isolated height −7 belongs. A one-member set can be written [−7, −7], and union must preserve all missing heights around it.