Question

question 10
if the function $k(x)=\sqrt{16 - x^2}$ is defined on $0\leq x\leq4$, which of the following expressions best defines $k^{-1}(x)$?
$\bigcirc$ $k^{-1}(x)=16 - x^2$ for $0\leq x\leq4$
$\bigcirc$ $k^{-1}(x)=\sqrt{16 - x^2}$ for $0\leq x\leq4$
$\bigcirc$ $k^{-1}(x)=\sqrt{16 - x}$ for $-4\leq x\leq0$
$\bigcirc$ $k^{-1}(x)=4 - x$ for $-4\leq x\leq0$

Explanation:

Step1: Set $y = k(x)$

$y = \sqrt{16 - x^2}$ where $0 \leq x \leq 4$

Step2: Swap $x$ and $y$

$x = \sqrt{16 - y^2}$

Step3: Square both sides

$x^2 = 16 - y^2$

Step4: Solve for $y^2$

$y^2 = 16 - x^2$

Step5: Find range of original function

For $0 \leq x \leq 4$, $0 \leq 16-x^2 \leq 16$, so $0 \leq y \leq 4$. Thus inverse domain is $0 \leq x \leq 4$.

Step6: Solve for $y$ (positive root)

$y = \sqrt{16 - x^2}$ with $0 \leq x \leq 4$

Answer:

$k^{-1}(x) = \sqrt{16 - x^2}$ for $0 \leq x \leq 4$