What is the extreme 3x^4-24x^3+48x^2 ?

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

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extreme 3x^4-24x^3+48x^2

extreme points

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)

(x - 1)^{4} \cdot (x - 3)^{2}

\frac{6 - 3 x}{x - 4}

f (x) = \frac{5}{x ^{2} - 16}

(0, 20), (13500, 4.5)

y = \frac{( x - 4 )}{( x + 1 )}

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