What is the extreme f(x)=2x^3+21x^2+36x ?

The extreme f(x)=2x^3+21x^2+36x is Maximum(-6,108),Minimum(-1,-17)

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extreme f(x)=2x^3+21x^2+36x

extreme points

f (x) = 2 x^{3} + 21 x^{2} + 36 x

Maximum (- 6, 108), Minimum (- 1, - 17)

Solution steps

f^{'} (x) = 6 x^{2} + 42 x + 36

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 6

into

2 x^{3} + 21 x^{2} + 36 x : 108

Maximum (- 6, 108)

Plug

x = - 1

into

2 x^{3} + 21 x^{2} + 36 x : - 17

Minimum (- 1, - 17)

Maximum (- 6, 108), Minimum (- 1, - 17)

\frac{( x - 10 ) ^{3}}{10} + 5

f (x) = \frac{2 x}{x - 1}

f (x) = ln (2 + 3 + x^{2})

2 y + x - 6 = 0

f (x) = - 3 x - 1

What is the extreme f(x)=2x^3+21x^2+36x ?
The extreme f(x)=2x^3+21x^2+36x is Maximum(-6,108),Minimum(-1,-17)