Answer :
Answer:
[tex]x=-2.3094[/tex] is a relative maxima and [tex]x=2.3094[/tex] is a relative minima.
Step-by-step explanation:
We have been given a function [tex]f(x)=\frac{1}{8}x^3-2x[/tex]. 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:
[tex]f'(x)=\frac{d}{dx}(\frac{1}{8}x^3)-\frac{d}{dx}(2x)[/tex]
[tex]f'(x)=3*\frac{1}{8}x^{3-1}-2*(x^{1-1})[/tex]
[tex]f'(x)=\frac{3}{8}x^{2}-2*(x^0)[/tex]
[tex]f'(x)=\frac{3}{8}x^{2}-2*(1)[/tex]
[tex]f'(x)=\frac{3}{8}x^{2}-2[/tex]
Now, we will find the critical points by equating derivative to 0 as:
[tex]\frac{3}{8}x^{2}-2=0[/tex]
[tex]\frac{3}{8}x^{2}=2[/tex]
[tex]\frac{8}{3}*\frac{3}{8}x^{2}=\frac{8}{3}*2[/tex]
[tex]x^{2}=\frac{16}{3}[/tex]
[tex]x=\pm \sqrt{\frac{16}{3}}[/tex]
[tex]x=\pm 2.3094[/tex]
Noe, we will check on which intervals our given function is increasing or decreasing.
[tex]f'(-4)=\frac{3}{8}(-4)^{2}-2[/tex]
[tex]f'(-4)=\frac{3}{8}(16)-2[/tex]
[tex]f'(-4)=3*2-2[/tex]
[tex]f'(-4)=4[/tex]
[tex]f'(1)=\frac{3}{8}(1)^{2}-2[/tex]
[tex]f'(1)=\frac{3}{8}-2[/tex]
[tex]f'(1)=-1.625[/tex]
[tex]f'(4)=\frac{3}{8}(4)^{2}-2[/tex]
[tex]f'(4)=\frac{3}{8}(16)-2[/tex]
[tex]f'(4)=3*2-2[/tex]
[tex]f'(4)=4[/tex]
We know that when [tex]f'(x)>0[/tex], then f is increasing and when [tex]f'(x)<0[/tex], then f is decreasing.
Therefore, [tex]x=-2.3094[/tex] is a relative maxima and [tex]x=2.3094[/tex] is a relative minima.