Question

In exercise, find the relative extrema of the function. f(x) = 1/8x^3 - 2x

Answer 1

Answer:

$x=-2.3094$ is a relative maxima and $x=2.3094$ is a relative minima.

Step-by-step explanation:

We have been given a function $f(x)=\frac{1}{8}x^3-2x$. We are asked to find the relative extrema of the given function.

First of all, we will find first derivative of the given function as:

$f'(x)=\frac{d}{dx}(\frac{1}{8}x^3)-\frac{d}{dx}(2x)$

$f'(x)=3*\frac{1}{8}x^{3-1}-2*(x^{1-1})$

$f'(x)=\frac{3}{8}x^{2}-2*(x^0)$

$f'(x)=\frac{3}{8}x^{2}-2*(1)$

$f'(x)=\frac{3}{8}x^{2}-2$

Now, we will find the critical points by equating derivative to 0 as:

$\frac{3}{8}x^{2}-2=0$

$\frac{3}{8}x^{2}=2$

$\frac{8}{3}*\frac{3}{8}x^{2}=\frac{8}{3}*2$

$x^{2}=\frac{16}{3}$  

$x=\pm \sqrt{\frac{16}{3}}$    

$x=\pm 2.3094$    

Noe, we will check on which intervals our given function is increasing or decreasing.

$f'(-4)=\frac{3}{8}(-4)^{2}-2$

$f'(-4)=\frac{3}{8}(16)-2$

$f'(-4)=3*2-2$

$f'(-4)=4$

$f'(1)=\frac{3}{8}(1)^{2}-2$

$f'(1)=\frac{3}{8}-2$

$f'(1)=-1.625$

$f'(4)=\frac{3}{8}(4)^{2}-2$

$f'(4)=\frac{3}{8}(16)-2$

$f'(4)=3*2-2$

$f'(4)=4$

We know that when $f'(x)>0$, then f is increasing and when $f'(x)$, then f is decreasing.

Therefore, $x=-2.3094$ is a relative maxima and $x=2.3094$ is a relative minima.