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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

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

Maximum (0, 5)

Plug

x = 4

into

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

Minimum (4, - 27)

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

\frac{2 x ^{2} - 7 x + 3}{x ^{2} - 3 x - 2}

f (x) = \frac{4 y ^{2} + 16 y + 64}{y ^{3} - 6 y ^{2}}

f (x) = \frac{1}{\frac{20}{x - 5} - 2}

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

- (x - 7)^{2} + 4

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