find the absolute maximum and minimum values of the function, if they exist, over the indicated interval…

find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. also indicate the x - value at which each extremum occurs. f(x)=4x + 4/x; (0,∞) select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute minimum value is at x = and the absolute maximum value is at x = (use a comma to separate answers as needed.) b. the absolute maximum value is at x = and there is no absolute minimum. (use a comma to separate answers as needed.) c. the absolute minimum value is at x = and there is no absolute maximum. (use a comma to separate answers as needed.) d. there is no absolute minimum and there is no absolute maximum.

find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. also indicate the x - value at which each extremum occurs. f(x)=4x + 4/x; (0,∞) select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute minimum value is at x = and the absolute maximum value is at x = (use a comma to separate answers as needed.) b. the absolute maximum value is at x = and there is no absolute minimum. (use a comma to separate answers as needed.) c. the absolute minimum value is at x = and there is no absolute maximum. (use a comma to separate answers as needed.) d. there is no absolute minimum and there is no absolute maximum.

Answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=4x+\frac{4}{x}=4x + 4x^{-1}$ using the power - rule. The derivative $f^\prime(x)=4-4x^{-2}=4-\frac{4}{x^{2}}$.

Step2: Set the derivative equal to zero

Set $f^\prime(x) = 0$. So, $4-\frac{4}{x^{2}}=0$. Add $\frac{4}{x^{2}}$ to both sides: $4=\frac{4}{x^{2}}$. Cross - multiply to get $4x^{2}=4$, then $x^{2} = 1$, and $x=\pm1$. But the interval is $(0,\infty)$, so we only consider $x = 1$.

Step3: Use the second - derivative test

Find the second - derivative $f^{\prime\prime}(x)=\frac{8}{x^{3}}$. Evaluate $f^{\prime\prime}(1)=\frac{8}{1^{3}}=8>0$. Since $f^{\prime\prime}(1)>0$, the function has a local minimum at $x = 1$.

Step4: Find the minimum value

Substitute $x = 1$ into the original function $f(x)$. $f(1)=4\times1+\frac{4}{1}=8$. As $x\rightarrow0^{+}$, $f(x)=4x+\frac{4}{x}\rightarrow\infty$, and as $x\rightarrow\infty$, $f(x)=4x+\frac{4}{x}\rightarrow\infty$. So there is no absolute maximum on the interval $(0,\infty)$.

Answer:

C. The absolute minimum value is 8 at x = 1 and there is no absolute maximum.