What is the extreme f(x)=4x^3+48x^2+180x ?

The extreme f(x)=4x^3+48x^2+180x is Maximum(-5,-200),Minimum(-3,-216)

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extreme f(x)=4x^3+48x^2+180x

extreme points

f (x) = 4 x^{3} + 48 x^{2} + 180 x

Maximum (- 5, - 200), Minimum (- 3, - 216)

Solution steps

f^{'} (x) = 12 x^{2} + 96 x + 180

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 3,

Increasing

: - 3 < x < \infty

Plug

x = - 5

into

4 x^{3} + 48 x^{2} + 180 x : - 200

Maximum (- 5, - 200)

Plug

x = - 3

into

4 x^{3} + 48 x^{2} + 180 x : - 216

Minimum (- 3, - 216)

Maximum (- 5, - 200), Minimum (- 3, - 216)

f (x) = \frac{x ^{2} - 7 x + 26}{x - 5}

f (x) = x^{3} - x^{2} - 8 x + 4

f (x) = x^{3} - x^{2} - 8 x + 8

f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

f (x, y) = x^{3} + y^{3} - 24 x y

What is the extreme f(x)=4x^3+48x^2+180x ?
The extreme f(x)=4x^3+48x^2+180x is Maximum(-5,-200),Minimum(-3,-216)