Grade 12 · lesson 4 of 4

Restrict a Quadratic to Make an Inverse Function

Squaring on all real inputs loses sign information and is not one-to-one.

About 25 minutes, plus any extended practice

Supplied materials

Paper or a browser scratch area; a calculator is optional unless rounding is requested.
Supplied teaching case: The function f(x)=x²+2 has domain x≥0. Find its inverse and domain. Candidate answers: (A) f⁻¹(x)=x²−2, x≥0 (B) f⁻¹(x)=±√(x−2), all real x (C) f⁻¹(x)=√(x+2), x≥0 (D) f⁻¹(x)=√(x−2), x≥2.
Independent check inputs: The function f(x)=x²+3 has domain x≥0. Find its inverse and domain. Candidate answers: (A) f⁻¹(x)=x²−3, x≥0 (B) f⁻¹(x)=±√(x−3), all real x (C) f⁻¹(x)=√(x+3), x≥0 (D) f⁻¹(x)=√(x−3), x≥3.

Learn the idea

Function operations must respect input order and domain. A composition feeds one output into another function, while an inverse reverses a one-to-one relationship and exchanges the original domain and range. Squaring on all real inputs loses sign information and is not one-to-one. Restricting to nonnegative inputs makes the positive square root recover one unique input and supplies an inverse.

Here is the method on the supplied teaching case. The nonnegative domain makes f one-to-one. Solve y=x²+2 with x≥0: x=√(y−2). The range of f, [2,∞), becomes the inverse domain.

Do not return both square-root branches as a function output. The chosen original domain selects the branch, and the original range becomes the inverse’s domain; preserve that restriction explicitly.

Write or sketch the supplied teaching case, labeling its quantities and restrictions.
Apply the method shown in the worked solution, keeping each intermediate equality or justified inference.
Solve the independent check with its changed inputs before revealing feedback.
Complete the three practice cases; explain which mathematical rule rules out one distractor in each case.

Worked example

See the method in action

Teaching case: The function f(x)=x²+2 has domain x≥0. Find its inverse and domain. Candidate answers: (A) f⁻¹(x)=x²−2, x≥0 (B) f⁻¹(x)=±√(x−2), all real x (C) f⁻¹(x)=√(x+2), x≥0 (D) f⁻¹(x)=√(x−2), x≥2. Solution: The nonnegative domain makes f one-to-one. Solve y=x²+2 with x≥0: x=√(y−2). The range of f, [2,∞), becomes the inverse domain. The final answer is f⁻¹(x)=√(x−2), x≥2. Do not return both square-root branches as a function output. The chosen original domain selects the branch, and the original range becomes the inverse’s domain; preserve that restriction explicitly.

Your turn

Check your reasoning

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Helpful tips

Squaring on all real inputs loses sign information and is not one-to-one.
Do not return both square-root branches as a function output.

See the answer and explanation

f⁻¹(x)=√(x−3), x≥3

The nonnegative domain makes f one-to-one. Solve y=x²+3 with x≥0: x=√(y−3). The range of f, [3,∞), becomes the inverse domain.

Take it further

Changed-input task: The function f(x)=x²+9 has domain x≥0. Find its inverse and domain. Candidate answers: (A) f⁻¹(x)=x²−9, x≥0 (B) f⁻¹(x)=±√(x−9), all real x (C) f⁻¹(x)=√(x+9), x≥0 (D) f⁻¹(x)=√(x−9), x≥9. Show the mathematical reason, not only the selected answer. Expected result: f⁻¹(x)=√(x−9), x≥9. The nonnegative domain makes f one-to-one. Solve y=x²+9 with x≥0: x=√(y−9). The range of f, [9,∞), becomes the inverse domain.

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More practice: try new cases

Choose an answer for each problem. Explain your method on paper, then use the feedback to decide what to revisit.

1. The function f(x)=x²+4 has domain x≥0. Find its inverse and domain.f⁻¹(x)=√(x−4), x≥4f⁻¹(x)=±√(x−4), all real xf⁻¹(x)=√(x+4), x≥0f⁻¹(x)=x²−4, x≥0

Need a hint?

Squaring on all real inputs loses sign information and is not one-to-one.
Do not return both square-root branches as a function output.

2. The function f(x)=x²+5 has domain x≥0. Find its inverse and domain.f⁻¹(x)=±√(x−5), all real xf⁻¹(x)=√(x−5), x≥5f⁻¹(x)=√(x+5), x≥0f⁻¹(x)=x²−5, x≥0

Need a hint?

Squaring on all real inputs loses sign information and is not one-to-one.
Do not return both square-root branches as a function output.

3. The function f(x)=x²+6 has domain x≥0. Find its inverse and domain.f⁻¹(x)=±√(x−6), all real xf⁻¹(x)=√(x+6), x≥0f⁻¹(x)=√(x−6), x≥6f⁻¹(x)=x²−6, x≥0

Need a hint?

Squaring on all real inputs loses sign information and is not one-to-one.
Do not return both square-root branches as a function output.

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