24. -/2 points details my notes scalc9 3.3.024. find the local maximum and local minimum values of f using…

Explanation:

Step1: Find the first - derivative

Use the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = x^{2}$, $u'=2x$, $v=x - 7$, $v' = 1$. Then $f'(x)=\frac{2x(x - 7)-x^{2}\times1}{(x - 7)^{2}}=\frac{2x^{2}-14x - x^{2}}{(x - 7)^{2}}=\frac{x^{2}-14x}{(x - 7)^{2}}=\frac{x(x - 14)}{(x - 7)^{2}}$.

Step2: Find the critical points

Set $f'(x)=0$. Then $\frac{x(x - 14)}{(x - 7)^{2}}=0$. The numerator $x(x - 14)=0$ gives $x = 0$ and $x = 14$. The function is undefined at $x = 7$, but this is not in the domain of the original function's critical - point consideration since the domain of $f(x)$ is $x\neq7$.

Step3: First - derivative test

Consider the intervals $(-\infty,0)$, $(0,14)$ and $(14,\infty)$. For $x\in(-\infty,0)$, let $x=-1$. Then $f'(-1)=\frac{(-1)(-1 - 14)}{(-1 - 7)^{2}}=\frac{15}{64}>0$. For $x\in(0,14)$, let $x = 1$. Then $f'(1)=\frac{1(1 - 14)}{(1 - 7)^{2}}=\frac{-13}{36}<0$. For $x\in(14,\infty)$, let $x = 15$. Then $f'(15)=\frac{15(15 - 14)}{(15 - 7)^{2}}=\frac{15}{64}>0$. So $f(x)$ has a local maximum at $x = 0$ and a local minimum at $x = 14$.

Step4: Find the second - derivative

Use the quotient rule again on $f'(x)=\frac{x^{2}-14x}{(x - 7)^{2}}$. Let $u=x^{2}-14x$, $u'=2x - 14$, $v=(x - 7)^{2}$, $v'=2(x - 7)$. Then $f''(x)=\frac{(2x - 14)(x - 7)^{2}-(x^{2}-14x)\times2(x - 7)}{(x - 7)^{4}}=\frac{(x - 7)[(2x - 14)(x - 7)-2(x^{2}-14x)]}{(x - 7)^{4}}=\frac{2x^{2}-28x + 98-2x^{2}+28x}{(x - 7)^{3}}=\frac{98}{(x - 7)^{3}}$.

Step5: Evaluate the second - derivative at critical points

$f''(0)=\frac{98}{(0 - 7)^{3}}=\frac{98}{-343}<0$, so $x = 0$ is a local maximum. $f(0)=\frac{0^{2}}{0 - 7}=0$. $f''(14)=\frac{98}{(14 - 7)^{3}}=\frac{98}{343}>0$, so $x = 14$ is a local minimum. $f(14)=\frac{14^{2}}{14 - 7}=\frac{196}{7}=28$.

Answer:

local maximum value: $0$ local minimum value: $28$