show that f(x) = √(25 - x²) satisfies all assumptions of the mean value theorem on -5, 3, and find all…

show that f(x) = √(25 - x²) satisfies all assumptions of the mean value theorem on -5, 3, and find all values of c in (-5, 3) that satisfy the conclusion of the theorem. as your answer, please input the sum of all values c in decimal format with three significant digits after the decimal point.

show that f(x) = √(25 - x²) satisfies all assumptions of the mean value theorem on -5, 3, and find all values of c in (-5, 3) that satisfy the conclusion of the theorem. as your answer, please input the sum of all values c in decimal format with three significant digits after the decimal point.

Answer

Explanation:

Step1: Check assumptions

The function $f(x)=\sqrt{25 - x^{2}}$ is the upper - half of a circle centered at the origin with radius $r = 5$. It is continuous on the closed interval $[-5,3]$ since it is a composition of a square - root function and a polynomial function, and the expression inside the square - root $25 - x^{2}\geq0$ for $x\in[-5,5]$. The derivative $f^\prime(x)=\frac{-x}{\sqrt{25 - x^{2}}}$ exists for all $x\in(-5,5)$, so it is differentiable on the open interval $(-5,3)$.

Step2: Apply Mean - Value Theorem

The Mean - Value Theorem states that if $y = f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f^\prime(c)=\frac{f(b)-f(a)}{b - a}$. Here, $a=-5$, $b = 3$, $f(-5)=\sqrt{25-(-5)^{2}}=0$, $f(3)=\sqrt{25 - 3^{2}} = 4$. So, $\frac{f(3)-f(-5)}{3-(-5)}=\frac{4 - 0}{8}=\frac{1}{2}$.

Step3: Find $c$

Set $f^\prime(c)=\frac{-c}{\sqrt{25 - c^{2}}}=\frac{1}{2}$. Cross - multiply to get $-2c=\sqrt{25 - c^{2}}$. Square both sides: $4c^{2}=25 - c^{2}$. Rearrange to $5c^{2}=25$, so $c^{2}=5$, and $c=\pm\sqrt{5}$. But since we are looking for $c\in(-5,3)$, we take $c =-\sqrt{5}$.

Answer:

$-2.236$