What is the extreme f(x)=x^3-4x^2+5 ?

The extreme f(x)=x^3-4x^2+5 is Maximum(0,5),Minimum(8/3 ,-121/27)

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extreme f(x)=x^3-4x^2+5

extreme points

f (x) = x^{3} - 4 x^{2} + 5

Maximum (0, 5), Minimum (\frac{8}{3}, - \frac{121}{27})

Solution steps

f^{'} (x) = 3 x^{2} - 8 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{8}{3},

Increasing

: \frac{8}{3} < x < \infty

Plug

x = 0

into

x^{3} - 4 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = \frac{8}{3}

into

x^{3} - 4 x^{2} + 5 : - \frac{121}{27}

Minimum (\frac{8}{3}, - \frac{121}{27})

Maximum (0, 5), Minimum (\frac{8}{3}, - \frac{121}{27})

f (x) = \frac{3 x ^{2} - 6 x}{x ^{2} - 25}

3 = 3 (2) - 12

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

f (x) = 5^{x}

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

What is the extreme f(x)=x^3-4x^2+5 ?
The extreme f(x)=x^3-4x^2+5 is Maximum(0,5),Minimum(8/3 ,-121/27)