What is the extreme f(x)=(x-2)^3(3x+4)^2 ?

The extreme f(x)=(x-2)^3(3x+4)^2 is Maximum(-4/3 ,0),Minimum(0,-128),Saddle(2,0)

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extreme f(x)=(x-2)^3(3x+4)^2

extreme

Maximum (- \frac{4}{3}, 0), Minimum (0, - 128), Saddle (2, 0)

Solution steps

f^{'} (x) = 15 x (x - 2)^{2} (3 x + 4)

Find intervals:

Increasing

: - \infty < x < - \frac{4}{3},

Decreasing

: - \frac{4}{3} < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - \frac{4}{3}

into

(x - 2)^{3} (3 x + 4)^{2} : 0

Maximum (- \frac{4}{3}, 0)

Plug

x = 0

into

(x - 2)^{3} (3 x + 4)^{2} : - 128

Minimum (0, - 128)

Plug

x = 2

into

(x - 2)^{3} (3 x + 4)^{2} : 0

Saddle (2, 0)

Maximum (- \frac{4}{3}, 0), Minimum (0, - 128), Saddle (2, 0)

What is the extreme f(x)=(x-2)^3(3x+4)^2 ?
The extreme f(x)=(x-2)^3(3x+4)^2 is Maximum(-4/3 ,0),Minimum(0,-128),Saddle(2,0)