Task Sets: Restrict domain for invertibility - Common Core Math - HS Functions

A

Recall / Warm-Up

1.

Which of the following functions fails the Horizontal Line Test on its full domain?

A.

$f(x) = 2x + 3$

B.

$f(x) = x^2$

C.

$f(x) = \sqrt{x}$

D.

$f(x) = \frac{1}{x}$

2.

When you restrict the domain of $f(x) = x^2$ to $x \geq 0$ , what changes and what stays the same?

A.

Both the rule $x^2$ and the allowed inputs change.

B.

Only the set of allowed inputs changes; the rule $x^2$ stays the same.

C.

The rule becomes $|x|$ to enforce non-negativity.

D.

The range automatically becomes all real numbers.

3.

The function $f(x) = (x - 3)^2$ has its vertex at $x = 3$ . Which domain restriction makes $f$ one-to-one?

A.

$x \geq 0$

B.

$x \geq 3$

C.

$x \leq 0$

D.

$x \geq -3$

B

Fluency Practice

1.

$f(x) = x^2$ with domain $\mathbb{R}$ has no inverse. If we restrict to $x \geq 0$ , the inverse is $f^{-1}(x) = \sqrt{x}$ . Why does the restriction fix the problem?

A.

On $x \geq 0$ , no two inputs share the same output, so the inverse can map each output back to a unique input.

B.

On $x \geq 0$ , the function becomes linear and therefore invertible.

C.

Restricting to $x \geq 0$ changes the formula to $|x|$ , which is always positive.

D.

Restriction makes the range smaller, which forces the inverse to exist.

2.

The function $f(x) = (x - 4)^2$ has its vertex at $x = 4$ . To restrict the domain so that $f$ is one-to-one on the right branch, enter the smallest value of $x$ allowed (as a number).

3.

$f(x) = x^2$ is restricted to $x \geq 0$ . What is $f^{-1}(x)$ ?

A.

$f^{-1}(x) = x^2$

B.

$f^{-1}(x) = -\sqrt{x}$

C.

$f^{-1}(x) = \sqrt{x}$

D.

$f^{-1}(x) = \pm\sqrt{x}$

4.

$f(x) = x^2 + 2$ is restricted to $x \geq 0$ . The range of this restricted function is $[a, \infty)$ . What is $a$ ?

5.

$f(x) = x^2 + 2$ , restricted to $x \geq 0$ . Find $f^{-1}(x)$ by setting $y = x^2 + 2$ , swapping variables, and solving for $y$ (taking the positive root). Enter your answer as a function rule — type: sqrt(x-2).

C

Varied Practice

1.

The graph of $f(x) = (x - 3)^2$ is a parabola with vertex at $(3, 0)$ . To restrict the domain to the right branch, which interval should you use?

A.

$x \geq 0$

B.

$x \geq 3$

C.

$x \geq -3$

D.

$x \geq 1$

2.

$f(x) = x^2 - 4$ restricted to $x \geq 0$ . To find the inverse:
Swap variables: $x = y^2 - 4$ .
Solve for $y$ : $y^2 = \underline{\hspace{5em}}$ , so $y = \underline{\hspace{5em}}$ (positive root).
The domain of $f^{-1}$ is $[\underline{\hspace{5em}}, \infty)$ .

y-squared equals:

inverse formula:

domain lower bound:

3.

$f(x) = x^2$ is restricted to $x \leq 0$ (the left half). What is the inverse function?

A.

$f^{-1}(x) = \sqrt{x}$

B.

$f^{-1}(x) = -\sqrt{x}$

C.

$f^{-1}(x) = \pm\sqrt{x}$

D.

$f^{-1}(x) = x^2$

4.

$f(x) = (x+1)^2 - 3$ , restricted to $x \geq -1$ . The range of $f^{-1}$ equals the domain of the restricted $f$ . What is the smallest value in the range of $f^{-1}$ ?

5.

$f(x) = x^2$ can be restricted to $x \geq 0$ or to $x \leq 0$ . What are the two different inverse functions you get? Explain why they are different.

D

Word Problems

1.

A ball is launched upward. Its height in meters after $t$ seconds is $h(t) = -5t^2 + 20t$ for $0 \leq t \leq 4$ .

1.

Does $h$ have an inverse on its full domain $[0, 4]$ ? Explain why or why not.

A.

Yes — $h$ passes the Horizontal Line Test on $[0, 4]$ .

B.

No — the ball reaches the same height on the way up and on the way down, so two different $t$ -values can give the same $h$ .

C.

Yes — h is a polynomial, so it is always invertible.

D.

No — $h$ is not one-to-one because it is quadratic and thus not linear.

2.

To find a unique inverse for the upward portion of the flight, restrict to $0 \leq t \leq 2$ . On this restricted domain, what is the maximum height (the upper bound of the range of $h$ )?

2.

The area $A$ of a square with side length $s$ is $A(s) = s^2$ , defined for $s > 0$ .

Since $s > 0$ , $A$ is already one-to-one. What is $A^{-1}(49)$ ? (Give the side length when the area is 49.)

E

Error Analysis

1.

Student work on $f(x) = (x - 5)^2$ :

"To make $f$ invertible, I restrict to $x \geq 0$ because we always use $x \geq 0$ for parabolas."

What is wrong with the student's reasoning?

A.

The restriction $x \geq 0$ still contains the vertex at $x = 5$ in its interior, so $f$ is not one-to-one on $x \geq 0$ . The restriction should be $x \geq 5$ (or $x \leq 5$ ).

B.

The student should restrict to $x \geq -5$ instead of $x \geq 0$ .

C.

The student is correct — $x \geq 0$ always works for any parabola.

D.

The student should restrict to $x \geq 25$ because that is the minimum output.

2.

Student work:

" $f(x) = x^2$ can only be restricted to $x \geq 0$ . There is only one valid way to make it invertible."

Is the student correct? Explain.

F

Challenge / Extension

1.

$f(x) = (x - 2)^2 + 5$ , restricted to $x \geq 2$ . Find the domain of $f^{-1}$ (the lower bound $a$ of $[a, \infty)$ ).

2.