What is the extreme f(x)=x^3-9x^2+24 ?

The extreme f(x)=x^3-9x^2+24 is Maximum(0,24),Minimum(6,-84)

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extreme f(x)=x^3-9x^2+24

extreme points

f (x) = x^{3} - 9 x^{2} + 24

Maximum (0, 24), Minimum (6, - 84)

Solution steps

f^{'} (x) = 3 x^{2} - 18 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{3} - 9 x^{2} + 24 : 24

Maximum (0, 24)

Plug

x = 6

into

x^{3} - 9 x^{2} + 24 : - 84

Minimum (6, - 84)

Maximum (0, 24), Minimum (6, - 84)

\frac{4 x ^{2}}{x ^{2} - 9}

f (x, y) = e^{(x^{2} + 0.5 y^{2} - 5 x y - 3 x)}

f (x) = x^{2} - 15 x + 56

y = x^{4} - x^{3}

f (t) = t u (t - 3)

What is the extreme f(x)=x^3-9x^2+24 ?
The extreme f(x)=x^3-9x^2+24 is Maximum(0,24),Minimum(6,-84)