Use toolkit behavior to find a changed function's range

Explain it like I am five

Imagine relabeling the buttons on a familiar machine and then changing what comes out. If a square-root machine receives x + 4 instead of x, its inside first reaches zero when your button says −4. That is why its graph starts farther left. If you then double every output, zero stays zero while other heights double. You can understand these changes by doing the arithmetic in order. First check which original inputs survive the calculation. Then ask which final outputs can be reached. A familiar graph helps you see the behavior, but a few plotted points cannot guarantee that every height in an interval occurs.

The solid 2 curve receives zero at x = −4 and doubles each root output; both curves continue beyond the frame.

Reminder

Principal square root. $\sqrt{z^{2}}$ = |z|. You may replace it with z only when you already know z ≥ 0.
Solving backward. Undo outside operations first: if y = 2t + 3, subtract 3 and divide by 2 to get t = $y − 3 over 2$ .
Multiplying an inequality by a negative. A negative multiplier reverses order: t ≥ 0 implies −3t ≤ 0.

Why it works. Horizontal changes act on the number fed into the toolkit rule. Setting that inside number to zero locates a square-root start or a reciprocal input that would make its denominator zero. Outside multiplication and addition act on outputs instead. To prove a range, either solve backward for a valid input or explain why a continuous graph crosses every proposed height. Continuous means the graph has no jumps or breaks on the interval in question, so it cannot skip an intermediate height while passing from below it to above it.

RuleFind the domain from the inside arithmetic. Find the range from the final output, and show every claimed output is attained.
A function continuous on the whole real line, with arbitrarily negative and positive outputs, has range (−∞, ∞).

The same idea, five ways

Say it

Check what feeds the basic machine, then what changes its output.

Write it

Inside changes determine the permitted original inputs, while outside changes alter the attainable output heights.

In math

For $square root of x + 4$ , solve x + 4 ≥ 0 to get x ≥ −4
For 2 $square root of x + 4$ , y ≥ 0 and range [0, ∞)
Every y ≥ 0 comes from x = ( $y over 2$ $) squared$ − 4

Relabel a machine's buttons before its calculation, then change the result after the calculation.

See it

The original input and the number fed into the root are different quantities.

The same idea, other ways

Inside first

For $square root of x + 4$ , an original input of −4 feeds 0 into the square-root rule. An original input of −3 feeds 1. To reproduce the basic square-root point (1, 1), solve x + 4 = 1, giving x = −3. The point moved left by 4 because of this equation, not because a plus sign always moves every kind of quantity left.

Backward from a target

For 2 $square root of x + 4$ , suppose the desired output is 6. Divide the output by 2 to get root value 3. Square to get inside value 9. Subtract 4 to get input 5. Substitution returns 2 $square root of 9$ = 6. Doing the same steps with any y ≥ 0 gives an input for every nonnegative output.

Follow the arithmetic forward to check an input found by working backward.

Crossing a height

For a continuous graph, think of walking along a connected ramp. To get from a height below 10 to a height above 10, you must pass through height 10. A graph with a jump could skip that height, so the no-jump condition is essential.

.1Shifted reciprocal

Think of changing the number of groups on a sharing machine before it divides. For f(x) = $2 over x + 1$ , the denominator is the original input plus 1. At input −1, that number becomes zero and division fails. To give the basic reciprocal denominator a value of 1, you now enter 0, since 0 + 1 = 1. This algebra explains the leftward change before you use the shifted picture.

Formula: f(x) = $2 over x + 1$ . It takes the reciprocal of x + 1 and doubles it.
Shape: the reciprocal branches lie around the forbidden input x = −1; the output level zero is still never reached.
Domain: (−∞, −1) ∪ (−1, ∞), because x + 1 must not equal zero.
Range: (−∞, 0) ∪ (0, ∞). The nonzero numerator prevents output zero. For any y ≠ 0, choose x = $2 over y$ − 1; the denominator becomes $2 over y$ , so the output is y.

The branches approach the forbidden input −1 and omitted output zero, with no endpoint at the edge of the frame.

Reminder

Principal square root. $\sqrt{z^{2}}$ = |z|. You may replace it with z only when you already know z ≥ 0.
Solving backward. Undo outside operations first: if y = 2t + 3, subtract 3 and divide by 2 to get t = $y − 3 over 2$ .
Multiplying an inequality by a negative. A negative multiplier reverses order: t ≥ 0 implies −3t ≤ 0.

The same idea, five ways

Say it

Say add one, then divide two by the result.

Write it

The denominator fails at input −1, while no input produces output zero.

In math

f(x) = $2 over x + 1$
x ≠ −1
Domain: (−∞, −1) ∪ (−1, ∞)
y ≠ 0
Range: (−∞, 0) ∪ (0, ∞)

A sharing machine adds a group before division, moving the original input at which its divisor is zero.

See it

The branches approach the forbidden input −1 and omitted output zero, with no endpoint at the edge of the frame.

Worked exampleDomain and range of the given shifted reciprocal

Find the domain and range of f(x) = $2 over x + 1$ . You need every input for which division works and every output some such input can produce.

The branches approach the forbidden input −1 and omitted output zero, with no endpoint at the edge of the frame.

Domain: (−∞, −1) ∪ (−1, ∞) The endpoint symbols record which limits belong.

Range: (−∞, 0) ∪ (0, ∞) The endpoint symbols record which limits belong.

Require x + 1 ≠ 0. Subtract 1 to get x ≠ −1. At x = −1, the denominator is −1 + 1 = 0, confirming the excluded input. Domain: (−∞, −1) ∪ (−1, ∞).A denominator cannot equal zero; every other real input gives a defined fraction.
The output cannot be zero: if $2 over x + 1$ = 0, multiplying by the allowed denominator would give 2 = 0.A nonzero numerator cannot produce a zero fraction.
Take any proposed y ≠ 0 and solve y = $2 over x + 1$ : y(x + 1) = 2, then x + 1 = $2 over y$ , then x = $2 over y$ − 1.Multiplication undoes division by the denominator, division by the nonzero target y isolates x + 1, and subtraction undoes the inside addition.
This input is not −1 because $2 over y$ is not zero. Substituting it gives 2 ÷ $2 over y$ = y. Range: (−∞, 0) ∪ (0, ∞).Every proposed nonzero output has a valid input, so no other output is missing.

Answer

Domain: (−∞, −1) ∪ (−1, ∞)
Range: (−∞, 0) ∪ (0, ∞)

Check At x = 0 the output is 2; at x = −2 it is −2. Working backward from either output returns those inputs. At x = −1 division fails. The graph has positive and negative branches but no point of height zero, matching the arithmetic proof.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Exclude output −1 because input −1 is forbidden.

An excluded input is an x-value, not an output height. Output −1 is attained: f(−3) = = −1.

✓ Instead: Exclude −1 from the domain and 0 from the range.

Tips and tricks

Write the forbidden input and omitted output in separate columns on scratch paper.

.2Shifted and scaled square root

Imagine adding 4 to the area display before a square-root machine finds a side length, then doubling the returned length. The formula is f(x) = 2 $square root of x + 4$ . The original input may be negative as long as the inside area is not. At −4 the inside becomes zero, so the graph starts there. Doubling acts after the root, so it changes output heights and keeps their nonnegative sign.

Formula: f(x) = 2 $square root of x + 4$ . The principal square root is taken before multiplying by 2.
Shape: a square-root curve starting at (−4, 0), with heights doubled.
Domain: [−4, ∞), because x + 4 ≥ 0 gives x ≥ −4.
Range: [0, ∞). Principal square roots are nonnegative, and doubling preserves that sign. For any y ≥ 0, input x = ( $y over 2$ $) squared$ − 4 is valid and produces y.

The included start is (−4, 0); every later root value is doubled, and the curve has no final height.

Reminder

Principal square root. $\sqrt{z^{2}}$ = |z|. You may replace it with z only when you already know z ≥ 0.
Solving backward. Undo outside operations first: if y = 2t + 3, subtract 3 and divide by 2 to get t = $y − 3 over 2$ .
Multiplying an inequality by a negative. A negative multiplier reverses order: t ≥ 0 implies −3t ≤ 0.

The same idea, five ways

Say it

Say add four, take the square root, then double.

Write it

The root’s inside starts at zero when the original input is −4.

In math

f(x) = 2 $square root of x + 4$
x + 4 ≥ 0
x ≥ −4
Domain: [−4, ∞)
y ≥ 0
Range: [0, ∞)

Add four to the area display before finding a side, then double the displayed side.

See it

The included start is (−4, 0); every later root value is doubled, and the curve has no final height.

Worked exampleDomain and range of the given doubled square root

Find the domain and range of f(x) = 2 $square root of x + 4$ . You are deciding which original inputs give a real root and which final outputs are attainable.

The included start is (−4, 0); every later root value is doubled, and the curve has no final height.

Domain: [−4, ∞) The endpoint symbols record which limits belong.

Range: [0, ∞) The endpoint symbols record which limits belong.

Require x + 4 ≥ 0. Subtract 4 to obtain x ≥ −4. Domain: [−4, ∞).The radicand, the expression inside an even root, must be nonnegative. Zero is permitted here because the root is not a denominator.
At x = −4, the output is 2 $square root of 0$ = 0. All outputs are nonnegative.The principal square root returns a nonnegative number, and multiplying by positive 2 keeps that sign.
For any proposed y ≥ 0, solve y = 2 $square root of x + 4$ : divide by 2 to get $y over 2$ = $square root of x + 4$ , then square to get ( $y over 2$ $) squared$ = x + 4, then subtract 4 to get x = ( $y over 2$ $) squared$ − 4.Undo the outside multiplication before undoing the root and the inside addition. The chosen root value $y over 2$ is nonnegative.
The constructed input is at least −4. Substitution gives 2 $square root of ( y over 2 ) squared$ = 2 × $y over 2$ = y. Range: [0, ∞).The nonnegative target makes the principal-root check valid, and every such target is attained.

Answer

Domain: [−4, ∞)
Range: [0, ∞)

Check Input −4 gives output 0, and input 5 gives 2 = 6. Working backward from target 6 gives (6 ÷ 2 − 4 = 9 − 4 = 5. These checks confirm the start and the scaling; the input formula proves the entire range.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The domain is [0, ∞) because the formula contains a square root.

The root receives x + 4 rather than x. For example input −3 gives inside value 1 and output 2, which is real.

✓ Instead: Require x + 4 ≥ 0, so the domain starts at −4. The range starts at output zero.

Tips and tricks

The inside must be nonnegative, even when the original input is negative. Keep those two numbers separate.

.3Cubic minus a linear term

Picture walking along a ramp that bends but has no gaps or jumps. If one end goes farther and farther below you while the other goes farther and farther above you, the ramp must cross every possible height somewhere. The graph of f(x) = $x^{3}$ − x behaves that way. It is continuous, meaning its curve does not break or jump. Subtracting x changes the curve near zero, but it cannot stop the cubic term from reaching very large positive and negative outputs.

Formula: f(x) = $x^{3}$ − x, a polynomial, which is a sum or difference of constant multiples of whole-number powers of x.
Shape: a continuous curve with bends near zero; it is not merely a shifted graph of $x^{3}$ .
Domain: (−∞, ∞), because cubing and subtracting are defined for every real input.
Continuous here means no jumps or breaks. Powers, multiplication, and subtraction vary without jumps, so this polynomial graph is continuous.
Range: (−∞, ∞). For t ≥ 2, f(t) = t( $t^{2}$ − 1) ≥ 3t, which grows without bound; f(−t) = −f(t) falls without bound. A continuous curve between those outputs must cross every intermediate height.

The sample columns show both output signs, while continuity and behavior beyond the sample prove the full range.

Reminder

Principal square root. $\sqrt{z^{2}}$ = |z|. You may replace it with z only when you already know z ≥ 0.
Solving backward. Undo outside operations first: if y = 2t + 3, subtract 3 and divide by 2 to get t = $y − 3 over 2$ .
Multiplying an inequality by a negative. A negative multiplier reverses order: t ≥ 0 implies −3t ≤ 0.

The same idea, five ways

Say it

Say a connected curve cannot skip a height.

Write it

The continuous polynomial $x^{3}$ − x extends in both output directions, so it reaches every real output.

In math

f(x) = $x^{3}$ − x
For t ≥ 2: f(t) ≥ 3t
f(−t) = −f(t)
Domain and range: (−∞, ∞)

A connected ramp must pass through a middle height when you walk from below that height to above it.

See it

The sample columns show both output signs, while continuity and behavior beyond the sample prove the full range.

Worked exampleDomain and range of the given cubic difference

Find the domain and range of f(x) = $x^{3}$ − x. You are identifying every legal input and proving that every output you claim can actually occur.

The sample columns show both output signs, while continuity and behavior beyond the sample prove the full range.

Write domain (−∞, ∞).Cubing and subtracting use every real input, with no denominator or even-root restriction.
The graph is continuous on the whole real line, meaning no jumps or breaks.Powers and subtraction change gradually when the input changes, so this polynomial graph cannot jump over an intermediate output height.
For t ≥ 2, $t^{2}$ = t × t ≥ 2 × 2 = 4. Subtract 1 to get $t^{2}$ − 1 ≥ 3, then multiply by positive t: f(t) = t( $t^{2}$ − 1) ≥ 3t. Its positive outputs have no upper limit.Both positive factors are at least 2. Subtracting the same number and multiplying by a positive number preserve the comparison, and 3t can exceed any proposed height by choosing t large enough.
f(−t) = (−t $) cubed$ − (−t) = − $t^{3}$ + t = −( $t^{3}$ − t) = −f(t). Its negative outputs have no lower limit.Changing the input's sign reverses this formula's output, so every large positive height has an equally large negative partner.
Given any real target y, choose t greater than both 2 and |y|. Then f(−t) < y < f(t), and the continuous graph passes through height y between inputs −t and t. Range: (−∞, ∞).A graph with no jumps cannot go from below a target height to above it without attaining that height.

Answer

Domain: (−∞, ∞)
Range: (−∞, ∞)

Check Substitution gives f(−3) = −27 + 3 = −24 and f(3) = 27 − 3 = 24. The table also gives outputs −6, 0, and 6, agreeing with both signs and the opposite-input relationship. Samples confirm the arithmetic; continuity plus unbounded behavior establishes all real outputs.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: − x is the cubic moved down by x, so its range is read from one fixed vertical shift.

x is not one fixed number. The amount being subtracted changes with each input and alters the shape.

✓ Instead: Treat − x as a continuous polynomial with arbitrarily negative and positive outputs. Those facts establish its all-real range.

Tips and tricks

You do not need a formula solving the cubic for every y. A continuous crossing argument can prove every target is attained.

.4Absolute value after another function

Imagine a thermometer reporting positions on either side of zero. You now ask how far each reported position is from zero, rather than whether it is above or below. Applying absolute value to a function’s outputs makes that change. Keep the original input choices. Transform the entire old output set into distances. A negative endpoint moves to a positive distance, but the smaller resulting distance may occur inside the old interval at zero, rather than at either endpoint. A value such as −9 and a value such as 9 both have distance 9. To reach a target distance, it is enough for either of those old outputs to belong to the old range.

For g(x) = |f(x)| on f’s domain, turn every old range member into its nonnegative distance.
If the old range contains 0, the new range contains 0.
For a bounded interval, compare both endpoint distances for the output bound, then check whether an old output attains that bound. An excluded outer endpoint can give a bound that is approached but never reached.
Check all intermediate distances, rather than applying absolute value to only two endpoints.

Read the entire old range before turning it into distances.

Reminder

Absolute value as distance. |−8| = 8, |7| = 7, and |0| = 0.

The same idea, five ways

Say it

Say take the distance from zero of every old output.

Write it

Applying absolute value changes output heights to nonnegative distances while preserving the supplied domain.

In math

g(x) = |f(x)|
Old range: [−12, 7]
New output: y = |z| for an old output z
New range: {y | 0 ≤ y ≤ 12} = [0, 12]

Replace a signed thermometer reading with its distance from zero.

See it

Zero and twelve are both attained distances.

Worked exampleTake the distance of every output

A function f has range [−12, 7]. Find the range of g(x) = |f(x)| on the same domain. You are taking the distance from zero of every output f can produce.

These are all old output values, including zero.

Their distances cover every value from zero through twelve.

The given range contains every real output from −12 through 7, including 0.An interval includes the numbers between its limits, not only the endpoints.
Taking absolute value turns each negative output into its positive distance and leaves each nonnegative output unchanged.Absolute value is distance from zero, so its final outputs cannot be negative.
Output 0 is attained because f actually produces 0; output 12 is attained because f actually produces −12.A range consists of attained outputs, and both given values belong to f’s range.
For any desired distance y with 0 ≤ y ≤ 12, the old output −y lies in [−12, 0], so some allowed input produces it. At that input |f(x)| = |−y| = y.This supplies every intervening distance and proves there are no missing values in the claimed range.
No output distance can exceed 12: old outputs lie no farther than 12 from zero. Write [0, 12].The old left endpoint has distance 12 and the old right endpoint has distance 7, so 12 is the greatest possible distance.

Answer

Range of g: [0, 12].

Check For example, old output −8 gives new output 8 and old output 7 gives new output 7. Taking only the endpoints would suggest [7, 12] and miss the attained distance 0 and all smaller distances.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The old endpoints have distances 12 and 7, so the new range is [7, 12].

The old interval also contains zero and all numbers near it, so the new function attains distances smaller than 7.

✓ Instead: The full new range is [0, 12].

Tips and tricks

Mark zero in the old output interval if it belongs. Find the lower and upper distance bounds, then check which bounds are attained.

Strategy: step by step

1. Identify the toolkit calculation inside the formula and teach yourself what number is actually fed into it.
2. Apply all denominator and even-root restrictions to find the domain.
3. Track each operation on the output, including a negative multiplier or an added constant.
4. Prove the full range by constructing an input for any allowed output, or by using a continuous graph with heights on both sides of every target.
5. Confirm a few values with substitution or a picture, keeping the proof separate from the sample check.

Strategy

Find a changed function's domain and prove its range

Does a denominator become zero for any input?

YesFind those inputs, substitute to confirm the zero denominator, and exclude them.

NoCheck any even root's inside.

↓

Is an output multiplier negative?

YesReverse output order: a lower starting bound can become an upper bound.

NoKeep the output order, then account for any outside addition.

↓

Can you solve backward for a target output?

YesState its permitted values and show the constructed input gives it.

NoIf the graph is continuous across an interval and crosses both sides of each target, explain why it cannot skip that height.

↓

Does the final operation take absolute value?

YesTurn the whole old output set into distances. Check whether zero is reached, compare endpoint distances, and prove every claimed distance occurs.

NoTrack the actual final operation on the outputs.

Name the toolkit operation and the expression it receives.
Solve the arithmetic restrictions on the original input.
Track all operations performed on the toolkit output.
Find a valid input for each allowed target, or use a continuous crossing argument on an interval.
Substitute sample inputs to check the result.

Worked exampleA square root moved right and lifted

Find the domain and range of f(x) = $square root of x − 10$ + 3. Find an input that gives output 8. You are identifying permitted inputs, attainable final heights, and one input producing the specified height.

The start is (10, 3); the curve continues right and up beyond the displayed frame.

The inside must satisfy x − 10 ≥ 0, so add 10 to get x ≥ 10. Domain: [10, ∞).The square root accepts nonnegative inside values. The inside first equals zero at x = 10.
At x = 10, f(10) = $square root of 0$ + 3 = 3. Every output is at least 3.The principal square root is nonnegative, and the outside addition lifts that root value by 3.
For an arbitrary target y ≥ 3, subtract 3: $square root of x − 10$ = y − 3. The right side is nonnegative. Square and add 10 to obtain x = (y − 3 $) squared$ + 10.These operations undo the output addition and then the square root. The nonnegative condition keeps the principal-root convention.
This input is at least 10, and f((y − 3 $) squared$ + 10) = $square root of (y − 3 ) squared$ + 3 = y − 3 + 3 = y. Range: [3, ∞).Every proposed y ≥ 3 has a valid input, while no smaller output can occur.
For output 8, x = (8 − 3 $) squared$ + 10 = $5 squared$ + 10 = 25 + 10 = 35.Use the backward formula with the desired output y = 8.

Answer

Domain: [10, ∞)
Range: [3, ∞)
Output 8 occurs at input 35

Check Substitute the obtained input: f(35) = + 3 = + 3 = 5 + 3 = 8. The basic square-root start occurs at inside value zero, so the changed graph starts at (10, 3), independently confirming both closed endpoints.

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: add a height to a square

Find the domain and range of f(x) = $x^{2}$ + 8. You are checking the inputs and the possible square outputs after adding 8.

Adding 8 changes the minimum output but does not restrict inputs.

Domain: (−∞, ∞).Every real input can be squared and then have 8 added.
Since $x^{2}$ ≥ 0, f(x) ≥ 8; x = 0 attains output 8.Adding 8 moves the square's minimum from 0 to 8.
For any y ≥ 8, choose x = $square root of y − 8$ . Then $x^{2}$ + 8 = y − 8 + 8 = y. Range: [8, ∞).The target condition makes the root real and constructs an input for every claimed height.

Answer

Domain: (−∞, ∞)
Range: [8, ∞)

Check The quadratic graph's bottom is (0, 8). To check output 17, use input 3: + 8 = 9 + 8 = 17, matching x = = 3.

Rung 2Rung 2: move the square-root start

Find the domain and range of g(x) = $square root of x − 10$ . You are deciding which original inputs feed nonnegative numbers into the root and which root outputs occur.

The start moves to input 10, while the root still reaches zero and every positive output.

Require x − 10 ≥ 0; add 10 to get x ≥ 10. Domain: [10, ∞).The inside, rather than the original x alone, must be nonnegative.
g(10) = $square root of 0$ = 0, and every output is nonnegative.The root is principal and zero is included.
For any y ≥ 0, choose x = $y^{2}$ + 10. Then $square root of x − 10$ = $\sqrt{y^{2}}$ = y. Range: [0, ∞).The input is at least 10, and the nonnegative target passes the principal-root check.

Answer

Domain: [10, ∞)
Range: [0, ∞)

Check Target output 5 gives input + 10 = 35. Substitution gives = = 5. The graph starts at (10, 0), as the inside-zero equation predicts.

Rung 3Rung 3: reflect a square root and raise its start

Find the domain and range of h(x) = −3 $square root of x + 8$ + 2. You are tracking an inside addition, a negative output multiplier, and an outside addition.

The negative multiplier turns the start into a maximum output of 2, and the curve continues downward.

Require x + 8 ≥ 0; subtract 8 to get x ≥ −8. Domain: [−8, ∞).The square-root inside is nonnegative, first becoming zero at x = −8.
The principal root is at least 0. Multiplying it by −3 gives a number at most 0; adding 2 gives h(x) ≤ 2.A negative multiplier reverses the order of the output heights, so the starting height is now a maximum.
h(−8) = −3 $square root of 0$ + 2 = 2.The upper bound is attained and therefore included.
To find an input for any target y ≤ 2, start with y = −3 $square root of x + 8$ + 2. Subtract 2 from both sides: y − 2 = −3 $square root of x + 8$ .We are undoing the outside addition to find an input that reaches the target output, not solving a domain restriction.
Divide both sides by −3: $y − 2 over −3$ = $square root of x + 8$ . Rewrite the fraction as $2 − y over 3$ . This root value is nonnegative because y ≤ 2.Negative division changes both signs in the fraction. The target restriction is what lets a principal square root equal that value.
Square both sides: x + 8 = ( $2 − y over 3$ $) squared$ . Subtract 8: x = ( $2 − y over 3$ $) squared$ − 8.Squaring recovers the inside value from the nonnegative root value. Subtracting 8 then finds the original input.
This input is at least −8 and gives −3 × $2 − y over 3$ + 2 = −(2 − y) + 2 = y. Range: (−∞, 2].Every proposed output at or below 2 has a valid input, and no output exceeds 2.

Answer

Domain: [−8, ∞)
Range: (−∞, 2]

Check For target output −4, the input formula gives ( − 8 = − 8 = −4. Direct substitution gives −3 + 2 = −3 × 2 + 2 = −4. The curve drops from its included start (−8, 2).

Rung 4A decreasing inside and a reflected square root

Find the domain and range of K(x) = − $square root of 17 − x$ . You are checking which inputs make the root real and which final heights appear after reversing the output sign.

The inside 17 − x permits inputs left of 17, while the outside minus gives heights at or below zero.

Require 17 − x ≥ 0. Subtract 17: −x ≥ −17. Divide by −1 and reverse the comparison: x ≤ 17.The inside must be nonnegative; negative division reverses the order of the input condition.
At x = 17, K(17) = − $square root of 0$ = 0. For every valid input, K(x) ≤ 0.A principal root is nonnegative and its opposite is nonpositive. Nonpositive means negative or zero, written ≤ 0.
For any target y ≤ 0, solve y = − $square root of 17 − x$ . Multiply by −1: −y = $square root of 17 − x$ ; the right target is nonnegative. Square: $y^{2}$ = 17 − x. Subtract 17 and multiply by −1: x = 17 − $y^{2}$ .Undoing the final sign and the root constructs the original input that would produce the target output.
This input is at most 17. Substitution gives − $square root of 17 − (17 − y squared )$ = − $\sqrt{y^{2}}$ = −|y| = y, because y ≤ 0.The target sign makes |y| = −y, so every nonpositive output is attained.

Answer

Domain: (−∞, 17].
Range: (−∞, 0].

Check For target −3, the constructed input is 17 − 9 = 8. Then K(8) = − = −3. Input 18 fails because the radicand is −1; the endpoint 17 gives the included maximum height 0.

Rung 5Rung 4: a reciprocal with an omitted height of 3

Find the domain and range of p(x) = $5 over x − 7$ + 3. You are locating the forbidden denominator input and the final output height that no input reaches.

The excluded input is 7 and the excluded output is 3; the two exclusions come from different questions.

Require x − 7 ≠ 0, so x ≠ 7. At x = 7, the denominator is 7 − 7 = 0, confirming the excluded input. Domain: (−∞, 7) ∪ (7, ∞).Only input 7 makes the denominator zero.
The fraction cannot equal zero. Therefore p(x) cannot equal 3.If the final output were 3, subtracting 3 would force a fraction with nonzero numerator 5 to equal zero.
For any y ≠ 3, subtract 3 from y = $5 over x − 7$ + 3 and solve: x − 7 = $5 over y − 3$ , so x = 7 + $5 over y − 3$ .The target condition permits division by y − 3, and reversing the denominator calculation isolates x.
This input is not 7 and produces y. Range: (−∞, 3) ∪ (3, ∞).The added fraction is nonzero, and substitution returns 5 ÷ $5 over y − 3$ + 3 = y − 3 + 3 = y.

Answer

Domain: (−∞, 7) ∪ (7, ∞)
Range: (−∞, 3) ∪ (3, ∞)

Check Target 4 gives input 7 + 5 ÷ 1 = 12, and p(12) = 5 ÷ 5 + 3 = 4. Target 2 gives input 7 + 5 ÷ (−1) = 2, and p(2) = 5 ÷ (−5) + 3 = 2. Both sides of the missing height are attained.

Rung 6Rung 5: reciprocal squared below a new height

Find the domain and range of q(x) = − $4 over (x + 2 ) squared$ + 7. You are locating the forbidden squared denominator and tracking the negative fraction before adding 7.

Both branches lie below the omitted height 7, and input −2 is forbidden.

Find the input that makes the denominator zero: x + 2 = 0 gives x = −2. Substitution gives (−2 + 2 $) squared$ = $0 squared$ = 0, so exclude −2. Domain: (−∞, −2) ∪ (−2, ∞).A square equals zero only when its base equals zero, and division by that zero denominator would be undefined.
For every valid input, (x + 2 $) squared$ > 0, so − $4 over (x + 2 ) squared$ < 0. Adding 7 gives q(x) < 7.The fraction has a positive denominator and a negative numerator, and it never equals zero.
To find an input for any target y < 7, start with y = 7 − $4 over (x + 2 ) squared$ . Subtract 7: y − 7 = − $4 over (x + 2 ) squared$ . Multiply both sides by −1: 7 − y = $4 over (x + 2 ) squared$ .These steps isolate the fraction so we can find an input producing the target. The target condition gives 7 − y > 0.
Multiply both sides by the nonzero denominator (x + 2 $) squared$ : (7 − y)(x + 2 $) squared$ = 4. Divide both sides by the positive number 7 − y: (x + 2 $) squared$ = $4 over 7 − y$ .The first multiplication undoes division by the original denominator. The second division isolates the square; both divisions use nonzero quantities.
Choose x = −2 + $square root of 4 over 7 − y$ . This input is not −2, since the root is strictly positive.The positive number under the root gives a nonzero base whose square is the required denominator.
Substituting gives 7 − 4 ÷ $4 over 7 − y$ = 7 − (7 − y) = y. Range: (−∞, 7).Every output strictly below 7 has a permitted input, while 7 itself and greater outputs are impossible.

Answer

Domain: (−∞, −2) ∪ (−2, ∞)
Range: (−∞, 7)

Check For target 3, choose x = −2 + = −1; direct substitution gives q(−1) = −4 ÷ 1 + 7 = 3. For target 6, choose x = −2 + = 0; q(0) = −4 ÷ 4 + 7 = 6. The curve can approach height 7 but never reaches it.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The graph of starts at x = 4 because the formula has +4.

The input fed into the root is x + 4. It is zero when x = −4, not when x = 4.

✓ Instead: Solve x + 4 = 0 first. The graph starts at (−4, 0), and its domain is [−4, ∞).

✗ Not this: Any function with arbitrarily negative and positive outputs must attain every real output.

A graph with jumps could skip some intermediate heights. Reaching both directions is not enough by itself.

✓ Instead: For − x, the graph is continuous, with no jumps or breaks, and extends to both extremes. Together those facts show that every intermediate output is attained.

Tips and tricks

For an inside change, solve an equation such as x + 4 = 0 before describing the graph's movement.
For a square-root range, check the starting height and whether the outside multiplier is positive or negative.
For a range proof, an input formula or a continuous crossing argument covers infinitely many outputs at once.

Trap. Listing outputs that seem possible without showing that an input produces them. Excluding zero, or observing a sign, may give a restriction but does not by itself prove the whole range.