the function $f(x)=7x + 8x^{-1}$ has one local minimum and one local maximum. this function has a local…

Explanation:

Step1: Find the derivative of the function

The derivative of (f(x)=7x + 8x^{-1}) is (f^\prime(x)=7-8x^{-2}=7-\frac{8}{x^{2}}).

Step2: Set the derivative equal to zero and solve for (x)

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 second - derivative test

The second derivative (f^{\prime\prime}(x)=\frac{16}{x^{3}}).

• When (x =-\frac{2\sqrt{14}}{7}), (f^{\prime\prime}\left(-\frac{2\sqrt{14}}{7}\right)=\frac{16}{\left(-\frac{2\sqrt{14}}{7}\right)^{3}}<0). So (x =-\frac{2\sqrt{14}}{7}) is a local maximum. Substitute (x =-\frac{2\sqrt{14}}{7}) into (f(x)): [ \begin{align*} f\left(-\frac{2\sqrt{14}}{7}\right)&=7\times\left(-\frac{2\sqrt{14}}{7}\right)+8\times\left(-\frac{2\sqrt{14}}{7}\right)^{-1}\ &=- 2\sqrt{14}-2\sqrt{14}\ &=-4\sqrt{14} \end{align*} ]
• When (x=\frac{2\sqrt{14}}{7}), (f^{\prime\prime}\left(\frac{2\sqrt{14}}{7}\right)=\frac{16}{\left(\frac{2\sqrt{14}}{7}\right)^{3}}>0). So (x=\frac{2\sqrt{14}}{7}) is a local minimum. Substitute (x = \frac{2\sqrt{14}}{7}) into (f(x)): [ \begin{align*} f\left(\frac{2\sqrt{14}}{7}\right)&=7\times\frac{2\sqrt{14}}{7}+8\times\left(\frac{2\sqrt{14}}{7}\right)^{-1}\ &=2\sqrt{14}+2\sqrt{14}\ &=4\sqrt{14} \end{align*} ]

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}).