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

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

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

extreme points

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

Minimum (0, 1), Maximum (2, 5)

Solution steps

f^{'} (x) = - 3 x^{2} + 6 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, 1)

Plug

x = 2

into

- x^{3} + 3 x^{2} + 1 : 5

Maximum (2, 5)

Minimum (0, 1), Maximum (2, 5)

2 + 9 x

f (x) = 6 x^{5}

f (x) = - 6 x

8 x e^{7 x}

f (x) = 2 + 2 x - 4

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