What is the extreme f(x)=x^3-4x^2+10 ?

The extreme f(x)=x^3-4x^2+10 is Maximum(0,10),Minimum(8/3 , 14/27)

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extreme f(x)=x^3-4x^2+10

extreme points

f (x) = x^{3} - 4 x^{2} + 10

Maximum (0, 10), Minimum (\frac{8}{3}, \frac{14}{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} + 10 : 10

Maximum (0, 10)

Plug

x = \frac{8}{3}

into

x^{3} - 4 x^{2} + 10 : \frac{14}{27}

Minimum (\frac{8}{3}, \frac{14}{27})

Maximum (0, 10), Minimum (\frac{8}{3}, \frac{14}{27})

f (x) = 19 x^{2} + 4 y = 76

(5, 2), (2, - 1)

f (x) = - 1 - x

(0, 1), (9, 10)

f (x) = - 2 x + 7

What is the extreme f(x)=x^3-4x^2+10 ?
The extreme f(x)=x^3-4x^2+10 is Maximum(0,10),Minimum(8/3 , 14/27)