What is the extreme f(x)=-x^3+3x^2-5 ?

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

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extreme f(x)=-x^3+3x^2-5

extreme points

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

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

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} - 5 : - 5

Minimum (0, - 5)

Plug

x = 2

into

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

Maximum (2, - 1)

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

f (x) = - x^{2} + 9

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

sin^{2} (θ)

\frac{- 2 - 5 x}{3 x - 1}

f (x) = \frac{x - 4}{x + 3}

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