show that (f(x)=\frac{9 - x^{2}}{4x}) satisfies all assumptions of the mean value theorem on (-4,-1), and…

Explanation:

Step1: Check continuity and differentiability

The function $f(x)=\frac{9 - x^{2}}{4x}=\frac{9}{4x}-\frac{x}{4}$ is a rational - function. It is continuous on the open interval $(-4,-1)$ and differentiable on the open interval $(-4,-1)$ since the denominator is non - zero on this interval. Also, it is defined on the closed interval $[-4,-1]$ (except at $x = 0$ which is outside $[-4,-1]$).

Step2: Calculate $f(-4)$ and $f(-1)$

$f(-4)=\frac{9-(-4)^{2}}{4\times(-4)}=\frac{9 - 16}{-16}=\frac{-7}{-16}=\frac{7}{16}$ $f(-1)=\frac{9-(-1)^{2}}{4\times(-1)}=\frac{9 - 1}{-4}=\frac{8}{-4}=-2$

Step3: Find the derivative of $f(x)$

Using the quotient rule, if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 9 - x^{2}$, $u^\prime=-2x$, $v = 4x$, $v^\prime = 4$. So $f^\prime(x)=\frac{-2x\times(4x)-(9 - x^{2})\times4}{(4x)^{2}}=\frac{-8x^{2}-36 + 4x^{2}}{16x^{2}}=\frac{-4x^{2}-36}{16x^{2}}=\frac{-x^{2}-9}{4x^{2}}$. By the Mean - Value Theorem, $f^\prime(c)=\frac{f(-1)-f(-4)}{-1-(-4)}$. $\frac{f(-1)-f(-4)}{-1 - (-4)}=\frac{-2-\frac{7}{16}}{3}=\frac{\frac{-32 - 7}{16}}{3}=\frac{-\frac{39}{16}}{3}=-\frac{13}{16}$

Step4: Solve for $c$

Set $f^\prime(c)=-\frac{13}{16}$, so $\frac{-c^{2}-9}{4c^{2}}=-\frac{13}{16}$. Cross - multiply: $16(-c^{2}-9)=-13\times4c^{2}$. Expand: $-16c^{2}-144=-52c^{2}$. Move terms involving $c^{2}$ to one side: $-16c^{2}+52c^{2}=144$. $36c^{2}=144$. $c^{2}=4$. $c=\pm2$. Since $c\in(-4,-1)$, $c=-2$.

Answer:

$-2.000$