Question

What are the local extrema of `f(x)= ((x-2)(x-4)^3)/(x^2-2)`?

Answer 1

`X=-5`

Explanation:

`f(x)=[(x-2)(x-4)^3]/(x^2-2)`
`x^2-2=(x+2)(x-2)`
So the function will became:
`f(x)=[(x-4)^3]/(x+2)`
Now
`f'(x)=d/dx[(x-4)^3]/(x+2)`
`f'(x)=[3(x+2)(x-4)^2-(x-4)^3]/(x+2)^2`
For local extremum point
`f'(x)=0`
So
`[3(x+2)(x-4)^2-(x-4)^3]/(x+2)^2=0`
`[3(x+2)(x-4)^2-(x-4)^3]=0`
`3(x+2)(x-4)^2=(x-4)^3`
`3x+6=x-4`
`2x=-10`
`x=-5`