find the locations of the absolute extrema of the function on the given interval. f(x)=25x/(x² + 1);-2,2 the…

Explanation:

Step1: Find the derivative

Use the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 25x$, $u'=25$, $v=x^{2}+1$, $v' = 2x$. Then $f'(x)=\frac{25(x^{2}+1)-25x(2x)}{(x^{2}+1)^{2}}=\frac{25x^{2}+25 - 50x^{2}}{(x^{2}+1)^{2}}=\frac{25 - 25x^{2}}{(x^{2}+1)^{2}}$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$, so $\frac{25 - 25x^{2}}{(x^{2}+1)^{2}}=0$. Since $(x^{2}+1)^{2}>0$ for all real $x$, we solve $25 - 25x^{2}=0$. Then $25(1 - x^{2})=0$, which gives $x=\pm1$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x=-2,x = - 1,x = 1,x = 2$. $f(-2)=\frac{25\times(-2)}{(-2)^{2}+1}=\frac{-50}{5}=-10$. $f(-1)=\frac{25\times(-1)}{(-1)^{2}+1}=-\frac{25}{2}$. $f(1)=\frac{25\times1}{1^{2}+1}=\frac{25}{2}$. $f(2)=\frac{25\times2}{2^{2}+1}=\frac{50}{5}=10$.

Answer:

The absolute maximum occurs at $x = 1$. The absolute minimum occurs at $x=-1$.