Question

first derivative test: problem 2
(1 point)
the function ( f(x)=7x + 8x^{-1} ) has one local minimum and one local maximum.
this function has a local maximum at ( x= ) with value
and a local minimum at ( x= ) with value
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Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=7x + 8x^{-1}\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=7-8x^{-2}=7-\frac{8}{x^{2}}\).

Step2: Set the derivative equal to zero to find critical points

Set \(f^\prime(x) = 0\), so \(7-\frac{8}{x^{2}}=0\).

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Step3: Use the first - derivative test

Choose test intervals. Let's consider the intervals \((-\infty,-\frac{2\sqrt{14}}{7})\), \((-\frac{2\sqrt{14}}{7},\frac{2\sqrt{14}}{7})\) and \((\frac{2\sqrt{14}}{7},\infty)\).

• For \(x=-2\) (in the interval \((-\infty,-\frac{2\sqrt{14}}{7})\)), \(f^\prime(-2)=7-\frac{8}{4}=7 - 2=5>0\).
• For \(x = 1\) (in the interval \((-\frac{2\sqrt{14}}{7},\frac{2\sqrt{14}}{7})\)), \(f^\prime(1)=7 - 8=-1<0\).
• For \(x = 2\) (in the interval \((\frac{2\sqrt{14}}{7},\infty)\)), \(f^\prime(2)=7-\frac{8}{4}=7 - 2=5>0\).

Since the function changes from increasing (\(f^\prime(x)>0\)) to decreasing (\(f^\prime(x)<0\)) at \(x=-\frac{2\sqrt{14}}{7}\), it is a local maximum.
Since the function changes from decreasing (\(f^\prime(x)<0\)) to increasing (\(f^\prime(x)>0\)) at \(x=\frac{2\sqrt{14}}{7}\), it is a local minimum.

Step4: Find the function values at critical points

• For \(x =-\frac{2\sqrt{14}}{7}\), \(f(-\frac{2\sqrt{14}}{7})=7\times(-\frac{2\sqrt{14}}{7})+8\times(-\frac{7}{2\sqrt{14}})=-2\sqrt{14}-\frac{28}{\sqrt{14}}=-2\sqrt{14}-2\sqrt{14}=-4\sqrt{14}\).
• For \(x=\frac{2\sqrt{14}}{7}\), \(f(\frac{2\sqrt{14}}{7})=7\times\frac{2\sqrt{14}}{7}+8\times\frac{7}{2\sqrt{14}}=2\sqrt{14}+\frac{28}{\sqrt{14}}=2\sqrt{14}+2\sqrt{14}=4\sqrt{14}\).

Answer:

The function has a local maximum at \(x =-\frac{2\sqrt{14}}{7}\) with value \(-4\sqrt{14}\) and a local minimum at \(x=\frac{2\sqrt{14}}{7}\) with value \(4\sqrt{14}\).