Back to Tutor Intake Assessment: Solve for inverse function

HSF.BF.B.4.a Tutor Intake -- Finding and Using Inverse Functions

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Grade 9·11 problems·~14 min·Common Core Math - HS Functions·standard·hsf-bf-b-4a

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Concepts

The inverse function reverses inputs and outputs. If $f(a) = b$ , then
$f^{-1}(b) = \underline{\hspace{5em}}$ .

missing value:

A student wants to find $f^{-1}(x)$ for $f(x) = 5x - 3$ . Which sequence
of steps is correct?

Write $y = 5x - 3$ ; solve for $x$ to get $x = \frac{y+3}{5}$ ; write
$f^{-1}(x) = \frac{y+3}{5}$ .

Write $y = 5x - 3$ ; swap $x$ and $y$ to get $x = 5y - 3$ ; solve for $y$
to get $y = \frac{x+3}{5}$ ; write $f^{-1}(x) = \frac{x+3}{5}$ .

Write $y = 5x - 3$ ; swap $x$ and $y$ to get $x = 5y - 3$ ; solve for $y$
to get $y = \frac{x-3}{5}$ ; write $f^{-1}(x) = \frac{x-3}{5}$ .

Write $y = 5x - 3$ ; replace $x$ with $-x$ to get $y = -5x - 3$ ; write
$f^{-1}(x) = -5x - 3$ .

For $f(x) = x^2$ , why can we not find an inverse function on the domain
of all real numbers?

Because $x^2$ is always positive, the inverse would have to take negative
inputs, which is not allowed.

Because $x^2$ is not one-to-one: both $x = 3$ and $x = -3$ give $f(x) = 9$ ,
so the inverse would not be a function.

Because the inverse of squaring is the square root, and square roots are
not defined for negative numbers.

Because $f(x) = x^2$ has no $x$ -intercept other than zero.

Procedures

Let $f(x) = 4x + 7$ . Solve $f(x) = 19$ for $x$ .

Enter the value of $x$ .

Using the swap-and-solve method, which expression gives $f^{-1}(x)$ for
$f(x) = \frac{x - 4}{3}$ ?

$\frac{3}{x - 4}$

$3x + 4$

$\frac{x + 4}{3}$

$3x - 4$

Let $f(x) = 2x^3$ . The inverse function is $f^{-1}(x) = \sqrt[3]{\frac{x}{2}}$ .

Use the inverse to evaluate: $f^{-1}(16)$ .

Enter your answer. (If the answer is not an integer, enter the exact decimal
or fraction — for example, enter 2 for $\sqrt[3]{8}$ .)

Using the swap-and-solve method on $f(x) = 2x^3$ , which expression gives
$f^{-1}(x)$ ?

$2\sqrt[3]{x}$

$\sqrt[3]{2x}$

$\sqrt[3]{\frac{x}{2}}$

$\frac{x^3}{2}$

Let $f(x) = 3x - 9$ . The inverse function is $f^{-1}(x) = \frac{x + 9}{3}$ .

Use the inverse to solve $f(x) = 6$ . That is, evaluate $f^{-1}(6)$ .

Enter the value of $x$ .

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