first derivative test: problem 2\n(1 point)\nthe function ( f(x)=7x + 8x^{-1} ) has one local minimum and…

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). [ \begin{align*} 7-\frac{8}{x^{2}}&=0\ 7x^{2}-8&=0\ x^{2}&=\frac{8}{7}\ x&=\pm\sqrt{\frac{8}{7}}=\pm\frac{2\sqrt{14}}{7} \end{align*} ]

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}).