What is the extreme f(x)=2x^3+3x^2+1 ?

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

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extreme f(x)=2x^3+3x^2+1

extreme points

f (x) = 2 x^{3} + 3 x^{2} + 1

Maximum (- 1, 2), Minimum (0, 1)

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} + 1 : 2

Maximum (- 1, 2)

Plug

x = 0

into

2 x^{3} + 3 x^{2} + 1 : 1

Minimum (0, 1)

Maximum (- 1, 2), Minimum (0, 1)

(2 x - x^{2}) e^{x}

f (x) = \frac{x + 3}{x - 3}

\frac{x ^{2} - 2 x - 3}{x}

f (x) = 3 + 8 + x

m = \frac{9}{2}

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