Question

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

Answer 1

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

Domain: $(-\infin, \infin)$

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

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

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

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

Find the critical number(s)

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

$$x_1=-1, x_2=1.4, x_3=3$$

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.

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

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

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

$$=(x+1)(10x^2-14x-30x+42)$$

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

$$+(x+1)(5x^2-15x+5x-15)$$

$$=(x+1)(20x^2-56x+20)$$

$$f''(-1)=0$$

$$f''(1.4)=-46.08<0$$

$$f''(3)=128>0$$

The function $f(x)$ has a local maximum at $x=1.4.$

The function $f(x)$ has a local minimum at $x=3.$

The function $f(x)$ has neither maximum nor minimum at $x=-1.$