What is the extreme f(x)=x^3-9x^2+4 ?

The extreme f(x)=x^3-9x^2+4 is Maximum(0,4),Minimum(6,-104)

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extreme f(x)=x^3-9x^2+4

extreme points

f (x) = x^{3} - 9 x^{2} + 4

Maximum (0, 4), Minimum (6, - 104)

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} + 4 : 4

Maximum (0, 4)

Plug

x = 6

into

x^{3} - 9 x^{2} + 4 : - 104

Minimum (6, - 104)

Maximum (0, 4), Minimum (6, - 104)

f (x) = \frac{1}{16 - x ^{2} + 4 y ^{2}}

f (x) = x^{3} (1 - x)^{2}

y = x^{\frac{2}{3}} (6 - x)^{\frac{1}{3}}

f (x) = (x^{2} - 4)^{7}

y = x^{3} - 27 x + 3, - 2 \leq x \leq 4

What is the extreme f(x)=x^3-9x^2+4 ?
The extreme f(x)=x^3-9x^2+4 is Maximum(0,4),Minimum(6,-104)