Question

1. $f(x)=\frac{x^{2}}{x + 2};-1,10$

Explanation:

Step1: Find the derivative

Use the quotient - rule. If $f(x)=\frac{u(x)}{v(x)}$, then $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2}$. Here, $u(x)=x^{2}$, $u^\prime(x) = 2x$, $v(x)=x + 2$, $v^\prime(x)=1$. So $f^\prime(x)=\frac{2x(x + 2)-x^{2}\times1}{(x + 2)^{2}}=\frac{2x^{2}+4x-x^{2}}{(x + 2)^{2}}=\frac{x^{2}+4x}{(x + 2)^{2}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$. So $\frac{x^{2}+4x}{(x + 2)^{2}}=0$. Since the denominator $(x + 2)^{2}
eq0$ for the domain we care about (we are looking at the interval $[-1,10]$), we set the numerator equal to 0. $x^{2}+4x=x(x + 4)=0$. The solutions are $x = 0$ and $x=-4$. But $x=-4$ is not in the interval $[-1,10]$, so the only critical point in the interval is $x = 0$.

Step3: Evaluate the function at the critical point and endpoints

Evaluate $f(x)$ at $x=-1$, $x = 0$, and $x = 10$.
When $x=-1$, $f(-1)=\frac{(-1)^{2}}{-1 + 2}=1$.
When $x = 0$, $f(0)=\frac{0^{2}}{0+2}=0$.
When $x = 10$, $f(10)=\frac{10^{2}}{10 + 2}=\frac{100}{12}=\frac{25}{3}$.

Answer:

The absolute minimum value of $f(x)$ on the interval $[-1,10]$ is $0$ (at $x = 0$) and the absolute maximum value is $\frac{25}{3}$ (at $x = 10$).