What is the extreme f(x)=3x^4-24x^3+48x^2 ?

The extreme f(x)=3x^4-24x^3+48x^2 is Minimum(0,0),Maximum(2,48),Minimum(4,0)

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extreme f(x)=3x^4-24x^3+48x^2

extreme points

f (x) = 3 x^{4} - 24 x^{3} + 48 x^{2}

Minimum (0, 0), Maximum (2, 48), Minimum (4, 0)

Solution steps

f^{'} (x) = 12 x^{3} - 72 x^{2} + 96 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

3 x^{4} - 24 x^{3} + 48 x^{2} : 0

Minimum (0, 0)

Plug

x = 2

into

3 x^{4} - 24 x^{3} + 48 x^{2} : 48

Maximum (2, 48)

Plug

x = 4

into

3 x^{4} - 24 x^{3} + 48 x^{2} : 0

Minimum (4, 0)

Minimum (0, 0), Maximum (2, 48), Minimum (4, 0)

f (x) = (x - 9)^{2} - 2

f (x) = \frac{x ^{2} - 9 x - 10}{x + 2}

\frac{x ^{2} - 3 x}{( x - 2 ) ^{2}}

\frac{1}{x ^{2} - 8 x}

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

What is the extreme f(x)=3x^4-24x^3+48x^2 ?
The extreme f(x)=3x^4-24x^3+48x^2 is Minimum(0,0),Maximum(2,48),Minimum(4,0)