find the absolute maximum and minimum values of the following function on the given interval. then graph the…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=-\frac{6}{x^{2}}=-6x^{-2}$ using the power - rule. The derivative $f^\prime(x)=(-6)\times(-2)x^{-3}=\frac{12}{x^{3}}$.

Step2: Find critical points

Set $f^\prime(x) = 0$. Since $\frac{12}{x^{3}}\neq0$ for any real $x$, there are no critical points in the domain of $f(x)$ where the derivative is zero. Also, the function $f(x)=-\frac{6}{x^{2}}$ is continuous on the closed interval $[1,2]$.

Step3: Evaluate the function at endpoints

Evaluate $f(x)$ at $x = 1$ and $x = 2$. When $x = 1$, $f(1)=-\frac{6}{1^{2}}=-6$. When $x = 2$, $f(2)=-\frac{6}{2^{2}}=-\frac{6}{4}=-\frac{3}{2}$.

Step4: Determine absolute extrema

Since $-6<-\frac{3}{2}$, the absolute maximum value of $f(x)$ on the interval $[1,2]$ is $-\frac{3}{2}$ which occurs at $x = 2$, and the absolute minimum value is $-6$ which occurs at $x = 1$.

Answer:

The absolute maximum value is $-\frac{3}{2}$ at $x = 2$, and the absolute minimum value is $-6$ at $x = 1$.