What is the extreme f(x)=4x^3-3x^2-6x+3 ?

The extreme f(x)=4x^3-3x^2-6x+3 is Maximum(-1/2 , 19/4),Minimum(1,-2)

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extreme f(x)=4x^3-3x^2-6x+3

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3

Maximum (- \frac{1}{2}, \frac{19}{4}), Minimum (1, - 2)

Solution steps

f^{'} (x) = 12 x^{2} - 6 x - 6

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = - \frac{1}{2}

into

4 x^{3} - 3 x^{2} - 6 x + 3 : \frac{19}{4}

Maximum (- \frac{1}{2}, \frac{19}{4})

Plug

x = 1

into

4 x^{3} - 3 x^{2} - 6 x + 3 : - 2

Minimum (1, - 2)

Maximum (- \frac{1}{2}, \frac{19}{4}), Minimum (1, - 2)

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

\frac{x ^{2}}{x - 5}

f (x, y) = x^{2} y + y^{3} - 12 y

f (x, y) = \frac{x y}{1 + e ^{x}}

f (x, y) = 5 + 9 - x^{2} - y^{2}

What is the extreme f(x)=4x^3-3x^2-6x+3 ?
The extreme f(x)=4x^3-3x^2-6x+3 is Maximum(-1/2 , 19/4),Minimum(1,-2)