given the function g(x) = (x^2 + 4)/8x, use the extreme value theorem to find the following. find the…

Explanation:

Step1: Rewrite the function

Rewrite $g(x)=\frac{x^{2}+4}{8x}=\frac{1}{8}x+\frac{1}{2x}$.

Step2: Find the derivative

Using the power - rule, $g^\prime(x)=\frac{1}{8}-\frac{1}{2x^{2}}$.

Step3: Set the derivative equal to zero

Set $\frac{1}{8}-\frac{1}{2x^{2}} = 0$. Then $\frac{1}{2x^{2}}=\frac{1}{8}$, which gives $x^{2}=4$, so $x = 2$ or $x=-2$. But since we are working on the interval $[1,10]$, we discard $x=-2$.

Step4: Evaluate the function at critical and end - points

Evaluate $g(x)$ at $x = 1$, $x=2$ and $x = 10$. $g(1)=\frac{1^{2}+4}{8\times1}=\frac{5}{8}$. $g(2)=\frac{2^{2}+4}{8\times2}=\frac{8}{16}=\frac{1}{2}$. $g(10)=\frac{10^{2}+4}{8\times10}=\frac{100 + 4}{80}=\frac{104}{80}=\frac{13}{10}$.

Answer:

The absolute maximum value is $\frac{13}{10}$. The absolute minimum value is $\frac{1}{2}$.