given the function g(x)=(x^2 + 4)/4x, use the extreme value theorem to find the following. find the absolute…

Explanation:

Step1: Rewrite the function

Rewrite $g(x)=\frac{x^{2}+4}{4x}$ as $g(x)=\frac{1}{4}x+\frac{1}{x}$.

Step2: Find the derivative

Using the power - rule, $g^\prime(x)=\frac{1}{4}-\frac{1}{x^{2}}$.

Step3: Set the derivative equal to zero

Set $\frac{1}{4}-\frac{1}{x^{2}} = 0$. Then $\frac{1}{x^{2}}=\frac{1}{4}$, so $x^{2}=4$ and $x = 2$ (we consider $x\in[1,10]$, and we ignore $x=-2$).

Step4: Evaluate the function at critical points and endpoints

Evaluate $g(x)$ at $x = 1$, $x=2$ and $x = 10$. When $x = 1$, $g(1)=\frac{1^{2}+4}{4\times1}=\frac{5}{4}$. When $x = 2$, $g(2)=\frac{2^{2}+4}{4\times2}=1$. When $x = 10$, $g(10)=\frac{10^{2}+4}{4\times10}=\frac{100 + 4}{40}=\frac{104}{40}=\frac{13}{5}$.

Answer:

The absolute maximum value is $\frac{13}{5}$ The absolute minimum value is $1$