Question

the function $f(x) = 8x + 4x^{-1}$ has one local minimum and one local maximum. this function has a local maximum at $x = \square$ and a local minimum at $x = \square$ question help: video message instructor

Explanation:

Step1: Find the first derivative

To find local extrema, we first find the derivative of \( f(x) = 8x + 4x^{-1} \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we have:
\( f^\prime(x)=\frac{d}{dx}(8x)+\frac{d}{dx}(4x^{-1}) = 8-4x^{-2} \)
Simplify \( f^\prime(x) \): \( f^\prime(x)=8-\frac{4}{x^{2}} \)

Step2: Find critical points

Set \( f^\prime(x) = 0 \) to find critical points:
\( 8-\frac{4}{x^{2}}=0 \)
Add \( \frac{4}{x^{2}} \) to both sides: \( 8=\frac{4}{x^{2}} \)
Cross - multiply: \( 8x^{2}=4 \)
Divide both sides by 8: \( x^{2}=\frac{4}{8}=\frac{1}{2} \)
Take square roots: \( x=\pm\sqrt{\frac{1}{2}}=\pm\frac{\sqrt{2}}{2}\approx\pm0.707 \)
We also note that \( x = 0 \) is not in the domain of \( f(x) \) (since \( f(x) \) has a term \( 4x^{-1}=\frac{4}{x} \))

Step3: Find the second derivative

To determine if these critical points are local maxima or minima, we find the second derivative. Differentiate \( f^\prime(x)=8 - 4x^{-2} \):
\( f^{\prime\prime}(x)=\frac{d}{dx}(8)-\frac{d}{dx}(4x^{-2})=0 + 8x^{-3}=\frac{8}{x^{3}} \)

Step4: Test critical points

• For \( x=\frac{\sqrt{2}}{2} \):

Substitute \( x = \frac{\sqrt{2}}{2} \) into \( f^{\prime\prime}(x) \). Since \( x=\frac{\sqrt{2}}{2}>0 \), \( f^{\prime\prime}(\frac{\sqrt{2}}{2})=\frac{8}{(\frac{\sqrt{2}}{2})^{3}} \).
\( (\frac{\sqrt{2}}{2})^{3}=\frac{(\sqrt{2})^{3}}{2^{3}}=\frac{2\sqrt{2}}{8}=\frac{\sqrt{2}}{4} \)
\( f^{\prime\prime}(\frac{\sqrt{2}}{2})=\frac{8}{\frac{\sqrt{2}}{4}}=8\times\frac{4}{\sqrt{2}}=\frac{32}{\sqrt{2}} = 16\sqrt{2}>0 \). So \( x = \frac{\sqrt{2}}{2} \) is a local minimum.

• For \( x=-\frac{\sqrt{2}}{2} \):

Substitute \( x=-\frac{\sqrt{2}}{2} \) into \( f^{\prime\prime}(x) \). Since \( x = -\frac{\sqrt{2}}{2}<0 \), \( f^{\prime\prime}(-\frac{\sqrt{2}}{2})=\frac{8}{(-\frac{\sqrt{2}}{2})^{3}} \).
\( (-\frac{\sqrt{2}}{2})^{3}=-\frac{(\sqrt{2})^{3}}{2^{3}}=-\frac{2\sqrt{2}}{8}=-\frac{\sqrt{2}}{4} \)
\( f^{\prime\prime}(-\frac{\sqrt{2}}{2})=\frac{8}{-\frac{\sqrt{2}}{4}}=8\times(-\frac{4}{\sqrt{2}})=-\frac{32}{\sqrt{2}}=-16\sqrt{2}<0 \). So \( x = -\frac{\sqrt{2}}{2} \) is a local maximum.

Answer:

The function has a local maximum at \( x=-\frac{\sqrt{2}}{2} \) (or \( x =-\frac{1}{\sqrt{2}} \)) and a local minimum at \( x=\frac{\sqrt{2}}{2} \) (or \( x=\frac{1}{\sqrt{2}} \))