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

The extreme f(x)=-2x^3+3x^2+36x is Minimum(-2,-44),Maximum(3,81)

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

extreme points

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

Minimum (- 2, - 44), Maximum (3, 81)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 2

into

- 2 x^{3} + 3 x^{2} + 36 x : - 44

Minimum (- 2, - 44)

Plug

x = 3

into

- 2 x^{3} + 3 x^{2} + 36 x : 81

Maximum (3, 81)

Minimum (- 2, - 44), Maximum (3, 81)

f (x) = - x^{3} + 27 x - 58

(7 + \frac{1}{x}) (\frac{1}{x})

f (x) = 5 x^{2} - \frac{12}{x + 3}

- sin (3 x)

\frac{x}{x - 3}

What is the extreme f(x)=-2x^3+3x^2+36x ?
The extreme f(x)=-2x^3+3x^2+36x is Minimum(-2,-44),Maximum(3,81)