hw 20 - absolute extrema section 3.6: problem 9\n(1 point)\nfind the extreme values of the function ( f ) on…

Explanation:

Step1: Find the derivative of (f(x))

The function is (f(x)=4x + \frac{5}{x}=4x+5x^{-1}). Using the power rule ((x^n)^\prime=nx^{n - 1}), the derivative (f^\prime(x)=4-5x^{-2}=4-\frac{5}{x^{2}}=\frac{4x^{2}-5}{x^{2}}).

Step2: Find the critical points

Set (f^\prime(x) = 0), so (\frac{4x^{2}-5}{x^{2}}=0). Since (x^{2}\gt0) for (x\in(0,\infty)), we solve (4x^{2}-5 = 0). (4x^{2}=5), then (x^{2}=\frac{5}{4}), and (x=\frac{\sqrt{5}}{2}) (because (x\gt0)).

Step3: Use the second - derivative test

Find the second - derivative (f^{\prime\prime}(x)). Differentiate (f^\prime(x)=4 - 5x^{-2}) with respect to (x). Using the power rule, (f^{\prime\prime}(x)=10x^{-3}=\frac{10}{x^{3}}). When (x = \frac{\sqrt{5}}{2}), (f^{\prime\prime}(\frac{\sqrt{5}}{2})=\frac{10}{(\frac{\sqrt{5}}{2})^{3}}\gt0). So (f(x)) has a local minimum at (x=\frac{\sqrt{5}}{2}).

Step4: Calculate the function value at the critical point

Substitute (x = \frac{\sqrt{5}}{2}) into (f(x)): (f(\frac{\sqrt{5}}{2})=4\times\frac{\sqrt{5}}{2}+\frac{5}{\frac{\sqrt{5}}{2}}=2\sqrt{5}+2\sqrt{5}=4\sqrt{5}). As (x\rightarrow0^{+}), (f(x)=4x+\frac{5}{x}\rightarrow\infty) (since (\lim_{x\rightarrow0^{+}}4x = 0) and (\lim_{x\rightarrow0^{+}}\frac{5}{x}=\infty)). As (x\rightarrow\infty), (f(x)=4x+\frac{5}{x}\rightarrow\infty) (since (\lim_{x\rightarrow\infty}4x=\infty) and (\lim_{x\rightarrow\infty}\frac{5}{x}=0)).

Answer:

Absolute minimum value: (4\sqrt{5}), located at (x=\frac{\sqrt{5}}{2}). Absolute maximum value: none, located at (x = ) none.