Question

section 2.8: optimization (homework) score: 10/120 answered: 1/12 question 2 0/10 pts 3 - 4 99 details score on last try: 0 of 10 pts. see details for more. at least one scored part is incorrect. jump to first changeable incorrect part. jump to last submitted part. next question get a similar question you can retry this question below the function ( f(x)=7x + 3x^{-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 question help: video

Explanation:

Step1: Find the derivative

The derivative of $f(x)=7x + 3x^{-1}$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=7-3x^{-2}=7-\frac{3}{x^{2}}$.

Step2: Set the derivative equal to zero

Set $f'(x) = 0$, so $7-\frac{3}{x^{2}}=0$. Then $\frac{3}{x^{2}}=7$, and $x^{2}=\frac{3}{7}$, which gives $x=\pm\sqrt{\frac{3}{7}}$.

Step3: Use the second - derivative test

Find the second - derivative $f''(x)=6x^{-3}=\frac{6}{x^{3}}$.
When $x =-\sqrt{\frac{3}{7}}$, $f''(-\sqrt{\frac{3}{7}})=\frac{6}{(-\sqrt{\frac{3}{7}})^{3}}<0$, so $f(x)$ has a local maximum at $x =-\sqrt{\frac{3}{7}}$.
When $x=\sqrt{\frac{3}{7}}$, $f''(\sqrt{\frac{3}{7}})=\frac{6}{(\sqrt{\frac{3}{7}})^{3}}>0$, so $f(x)$ has a local minimum at $x=\sqrt{\frac{3}{7}}$.

Step4: Find the function values

For the local maximum at $x =-\sqrt{\frac{3}{7}}$:
$f(-\sqrt{\frac{3}{7}})=7(-\sqrt{\frac{3}{7}})+3(-\sqrt{\frac{3}{7}})^{-1}=- \sqrt{21}-\sqrt{21}=-2\sqrt{21}$.
For the local minimum at $x=\sqrt{\frac{3}{7}}$:
$f(\sqrt{\frac{3}{7}})=7\sqrt{\frac{3}{7}}+3(\sqrt{\frac{3}{7}})^{-1}=\sqrt{21}+\sqrt{21}=2\sqrt{21}$.

Answer:

Local maximum at $x =-\sqrt{\frac{3}{7}}$ with value $-2\sqrt{21}$; Local minimum at $x=\sqrt{\frac{3}{7}}$ with value $2\sqrt{21}$