What is the extreme f(x)=(x+6)(x-3)^2 ?

The extreme f(x)=(x+6)(x-3)^2 is Maximum(-3,108),Minimum(3,0)

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extreme f(x)=(x+6)(x-3)^2

extreme points

f (x) = (x + 6) (x - 3)^{2}

Maximum (- 3, 108), Minimum (3, 0)

Solution steps

f^{'} (x) = 3 (x - 3) (x + 3)

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

(x + 6) (x - 3)^{2} : 108

Maximum (- 3, 108)

Plug

x = 3

into

(x + 6) (x - 3)^{2} : 0

Minimum (3, 0)

Maximum (- 3, 108), Minimum (3, 0)

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 10

f (x) = x^{2} - 2 x, 0 \leq x \leq 4

f (x) = x^{3} - x^{2} - 8 x + 8, - 2 \leq x \leq 0

f (x) = \frac{x ^{2} - 11 x + 54}{x - 9}

2 x y

What is the extreme f(x)=(x+6)(x-3)^2 ?
The extreme f(x)=(x+6)(x-3)^2 is Maximum(-3,108),Minimum(3,0)