What is the extreme f(x)=4x^3-12x^2+9x-1 ?

The extreme f(x)=4x^3-12x^2+9x-1 is Maximum(1/2 ,1),Minimum(3/2 ,-1)

extreme f(x)=4x^3-12x^2+9x-1

extreme points

f (x) = 4 x^{3} - 12 x^{2} + 9 x - 1

Maximum (\frac{1}{2}, 1), Minimum (\frac{3}{2}, - 1)

Solution steps

f^{'} (x) = 12 x^{2} - 24 x + 9

Find intervals:

Increasing

: - \infty < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = \frac{1}{2}

into

4 x^{3} - 12 x^{2} + 9 x - 1 : 1

Maximum (\frac{1}{2}, 1)

Plug

x = \frac{3}{2}

into

4 x^{3} - 12 x^{2} + 9 x - 1 : - 1

Minimum (\frac{3}{2}, - 1)

Maximum (\frac{1}{2}, 1), Minimum (\frac{3}{2}, - 1)

y = 2 x z

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

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

f (x) = 2 x^{5} - 6 x^{4}

f (x) = - (x^{2} + x)^{\frac{2}{3}}

What is the extreme f(x)=4x^3-12x^2+9x-1 ?
The extreme f(x)=4x^3-12x^2+9x-1 is Maximum(1/2 ,1),Minimum(3/2 ,-1)