What is the extreme f(x)=3x^3-24x^2+48x-6 ?

The extreme f(x)=3x^3-24x^2+48x-6 is Maximum(4/3 , 202/9),Minimum(4,-6)

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extreme f(x)=3x^3-24x^2+48x-6

extreme points

f (x) = 3 x^{3} - 24 x^{2} + 48 x - 6

Maximum (\frac{4}{3}, \frac{202}{9}), Minimum (4, - 6)

Solution steps

f^{'} (x) = 9 x^{2} - 48 x + 48

Find intervals:

Increasing

: - \infty < x < \frac{4}{3},

Decreasing

: \frac{4}{3} < x < 4,

Increasing

: 4 < x < \infty

Plug

x = \frac{4}{3}

into

3 x^{3} - 24 x^{2} + 48 x - 6 : \frac{202}{9}

Maximum (\frac{4}{3}, \frac{202}{9})

Plug

x = 4

into

3 x^{3} - 24 x^{2} + 48 x - 6 : - 6

Minimum (4, - 6)

Maximum (\frac{4}{3}, \frac{202}{9}), Minimum (4, - 6)

\frac{5}{x + 1}

f (x) = \frac{( x - 1 ) ^{2}}{x ^{2} - 1}

y = - (x - 1) (x + 2)^{3}

x^{4} - 4 x^{3} + 14

f (x) = \frac{2 x ^{2} + 18}{x}

What is the extreme f(x)=3x^3-24x^2+48x-6 ?
The extreme f(x)=3x^3-24x^2+48x-6 is Maximum(4/3 , 202/9),Minimum(4,-6)