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

The extreme f(x)=2x^3+3x^2+4 is Maximum(-1,5),Minimum(0,4)

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

extreme points

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

Maximum (- 1, 5), Minimum (0, 4)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

2 x^{3} + 3 x^{2} + 4 : 5

Maximum (- 1, 5)

Plug

x = 0

into

2 x^{3} + 3 x^{2} + 4 : 4

Minimum (0, 4)

Maximum (- 1, 5), Minimum (0, 4)

f (x) = x^{3} + 3 y^{3} + 3 x^{2} + 3 y^{2} + 24

f (x, y) = x + 4 y + \frac{2}{x y}

x e^{- x^{2}}

f (x) = 2 x^{3} + 12 x^{2} - 72 x + 11

f (x, y) = x^{4} + 6 y - 10

What is the extreme f(x)=2x^3+3x^2+4 ?
The extreme f(x)=2x^3+3x^2+4 is Maximum(-1,5),Minimum(0,4)