Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function. identify the points on the graph where the absolute extrema occur. f(x)= -\frac{6}{x^{2}}, 1\leq x\leq2 b. there is no absolute minimum. choose the correct graph of the function.

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 $\frac{12}{x^{3}}
eq0$ 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 $x = 1$.