What is the extreme f(x)=-x^3+1.5x^2+36x+2 ?

The extreme f(x)=-x^3+1.5x^2+36x+2 is Minimum(-3,-65.5),Maximum(4,106)

extreme f(x)=-x^3+1.5x^2+36x+2

extreme points

f (x) = - x^{3} + 1.5 x^{2} + 36 x + 2

Minimum (- 3, - 65.5), Maximum (4, 106)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 3

into

- x^{3} + 1.5 x^{2} + 36 x + 2 : - 65.5

Minimum (- 3, - 65.5)

Plug

x = 4

into

- x^{3} + 1.5 x^{2} + 36 x + 2 : 106

Maximum (4, 106)

Minimum (- 3, - 65.5), Maximum (4, 106)

f (x, y) = 2 x^{2} + y^{2} - 4 x + 6 y

y = x e^{- 2 x}

f (x) = x + 3 y^{2} = 25

f (x) = 110.2 x - 0.4 x^{2} - 20.2 - 2000

f (x) = \frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36

What is the extreme f(x)=-x^3+1.5x^2+36x+2 ?
The extreme f(x)=-x^3+1.5x^2+36x+2 is Minimum(-3,-65.5),Maximum(4,106)