Answer to Question #285797 in Real Analysis for Reens

Question

Examine the function ,f(x)=(x+1)^3(x-3)^2 for extreme values

Answer 1

$$f(x)=(x+1)^3(x-3)^2$$

Find the first derivative with respect to $x$

$$f'(x)=3(x+1)^2(x-3)^2+2(x+1)^3(x-3)$$

$$=(x+1)^2(x-3)(3x-9+2x+2)$$

$$=(x+1)^2(x-3)(5x-7)$$

Find the critical number(s)

$$f'(x)=0=>(x+1)^2(x-3)(5x-7)=0$$

Critical numbers: $-1, 1.4, 3.$

If $x<-1,f'(x)>0, f(x)$ increases.

If $-1<x<1.4,f'(x)>0, f(x)$ increases.

If $1.4<x<3,f'(x)<0, f(x)$ decreases.

If $x>3,f'(x)>0, f(x)$ increases.

$$f(-1)=(-1+1)^3(-1-3)^2=0$$

$$f(1.4)=(1.4+1)^3(1.4-3)^2=35.38944$$

$$f(3)=(3+1)^3(3-3)^2=0$$

The function $f$ has a local maximum with value of $35.38944$ at $x=1.4.$

The function $f$ has a local minimum with value of at $x=3.$

The function $f$ has neither a local maximum nor a local minimum at $x=-1.$