What is the extreme f(x)=-x^3+6x^2-5 ?

The extreme f(x)=-x^3+6x^2-5 is Minimum(0,-5),Maximum(4,27)

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extreme f(x)=-x^3+6x^2-5

extreme points

f (x) = - x^{3} + 6 x^{2} - 5

Minimum (0, - 5), Maximum (4, 27)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

- x^{3} + 6 x^{2} - 5 : - 5

Minimum (0, - 5)

Plug

x = 4

into

- x^{3} + 6 x^{2} - 5 : 27

Maximum (4, 27)

Minimum (0, - 5), Maximum (4, 27)

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

e^{7 x} + 7 e^{7 x} x

f (x) = \frac{x}{4 - x ^{2}}

y = x - 6

f (x) = lo g_{2} (x - 2)

What is the extreme f(x)=-x^3+6x^2-5 ?
The extreme f(x)=-x^3+6x^2-5 is Minimum(0,-5),Maximum(4,27)