Question

hw 20 - absolute extrema section 3.6: problem 9
(1 point)
find the extreme values of the function ( f ) on the interval ( (0, infty) ), and the ( x ) - value(s) at which they occur. if an extreme value does not exist, enter none for both the value and location.

( f(x)=4 x+\frac{5}{x} )
absolute minimum value: located at ( x= )
absolute maximum value: located at ( x= )

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 \(x
ightarrow0^{+}\), \(f(x)=4x+\frac{5}{x}
ightarrow\infty\) (since \(\lim_{x
ightarrow0^{+}}4x = 0\) and \(\lim_{x
ightarrow0^{+}}\frac{5}{x}=\infty\)).
As \(x
ightarrow\infty\), \(f(x)=4x+\frac{5}{x}
ightarrow\infty\) (since \(\lim_{x
ightarrow\infty}4x=\infty\) and \(\lim_{x
ightarrow\infty}\frac{5}{x}=0\)).

Answer:

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