Show that the absolute minimum of $f(x)=\sqrt{25-x^2}$ on [-5,5] is 0 and the absolute maximum is 5.

Here's a step-by-step explanation to show that the absolute minimum of $$\begin{gathered} f(x)= \\ sqrt25-x^2 \end{gathered}$$ on $[-5,5]$ is 0 and the absolute maximum is 5. Step 1: Find the first derivative of f(x) We can differentiate the function with respect to x. Since $$\begin{gathered} f(x)= \\ sqrt25-x^2 \end{gathered}$$, we can rewrite this as $$\begin{gathered} f(x)=(25-x^2{)}^{ \\ frac12} \end{gathered}$$ . Then, applying the chain rule to find the derivative: $$\begin{gathered} f^{\prime }(x)= \\ frac12(25-x^2{)}^{- \\ frac12}(-2x)=- \\ fracx \\ sqrt25-x^2 \end{gathered}$$ Step 2: Determine the critical points we must find where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined. The first derivative is undefined when the denominator is zero, which happens when x = ±5. So, let's find where the first derivative equals zero: $$\begin{gathered} frac-x \\ sqrt25-x^2=0 \end{gathered}$$ Multiplying both sides of the equation by $sqrt25-x^2$, we get: $-x=0$ So, we have one critical point at $x=0$. Step 3: Evaluate the function at critical points and endpoints We will now evaluate the function $f(x)$ at all critical points and the two endpoints (-5 and 5) since the function is continuous everywhere in $[-5,5]$. The two critical points are x = -5, 0, and 5. $$\begin{gathered} f(-5)= \\ sqrt25-(-5{)}^2= \\ sqrt0=0 \end{gathered}$$$$\begin{gathered} f(0)= \\ sqrt25-0^2= \\ sqrt25=5 \end{gathered}$$$$\begin{gathered} f(5)= \\ sqrt25-5^2= \\ sqrt0=0 \end{gathered}$$ Step 4: Determine the absolute minimum and maximum Comparing the values of the function, we see that: - Absolute minimum is 0, which occurs at x = -5 and x = 5. - Absolute maximum is 5, which occurs at x = 0. Hence, the absolute minimum of $f(x)$ on the interval $[-5,5]$ is 0, and the absolute maximum is 5.