What is the extreme f(x)=4x^3-3x^2-6x+17 ?

The extreme f(x)=4x^3-3x^2-6x+17 is Maximum(-1/2 , 75/4),Minimum(1,12)

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extreme f(x)=4x^3-3x^2-6x+17

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 17

Maximum (- \frac{1}{2}, \frac{75}{4}), Minimum (1, 12)

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 + 17 : \frac{75}{4}

Maximum (- \frac{1}{2}, \frac{75}{4})

Plug

x = 1

into

4 x^{3} - 3 x^{2} - 6 x + 17 : 12

Minimum (1, 12)

Maximum (- \frac{1}{2}, \frac{75}{4}), Minimum (1, 12)

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f (x) = \frac{x}{x + 1}

\frac{4}{x + 7}

f (t) = sin (2 t - \frac{π}{3}) + 1

What is the extreme f(x)=4x^3-3x^2-6x+17 ?
The extreme f(x)=4x^3-3x^2-6x+17 is Maximum(-1/2 , 75/4),Minimum(1,12)