Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function. g(x)=\sqrt{1 - x^{2}}, 0\leq x\leq1
a. the absolute minimum value 0 occurs at x = 1. (type exact answers, using radicals as needed. use a comma to separate answers as needed.)
b. there is no absolute minimum.
choose the correct graph of the function.

Explanation:

Step1: Evaluate function at endpoints

For $x = 0$, $g(0)=\sqrt{1 - 0^{2}}=1$. For $x = 1$, $g(1)=\sqrt{1 - 1^{2}}=0$.

Step2: Check for critical points

First, find the derivative of $g(x)$. Using the chain - rule, if $y=\sqrt{u}=u^{\frac{1}{2}}$ and $u = 1 - x^{2}$, then $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=-2x$. So $g^\prime(x)=\frac{-2x}{2\sqrt{1 - x^{2}}}=\frac{-x}{\sqrt{1 - x^{2}}}$. Set $g^\prime(x) = 0$, we get $x = 0$. The domain is $0\leq x\leq1$, and the derivative is not defined when $x = 1$.

Step3: Determine max and min

Comparing the function values at endpoints and critical points, the absolute maximum value of $g(x)$ on $[0,1]$ is $1$ at $x = 0$, and the absolute minimum value is $0$ at $x = 1$.

The function $y = \sqrt{1 - x^{2}}$ on $0\leq x\leq1$ is the upper - right quarter of the unit circle $x^{2}+y^{2}=1$. When $x = 0$, $y = 1$; when $x = 1$, $y = 0$.

Answer:

A. The absolute minimum value $0$ occurs at $x = 1$.
The correct graph is D (since it starts at $(0,1)$ and ends at $(1,0)$ for the portion of the function $y=\sqrt{1 - x^{2}}$ on the interval $[0,1]$).