What is the extreme f(x)=4x^3-3x^2-36x+17 ?

The extreme f(x)=4x^3-3x^2-36x+17 is Maximum(-3/2 , 203/4),Minimum(2,-35)

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extreme f(x)=4x^3-3x^2-36x+17

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 36 x + 17

Maximum (- \frac{3}{2}, \frac{203}{4}), Minimum (2, - 35)

Solution steps

f^{'} (x) = 12 x^{2} - 6 x - 36

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 2 < x < \infty

Plug

x = - \frac{3}{2}

into

4 x^{3} - 3 x^{2} - 36 x + 17 : \frac{203}{4}

Maximum (- \frac{3}{2}, \frac{203}{4})

Plug

x = 2

into

4 x^{3} - 3 x^{2} - 36 x + 17 : - 35

Minimum (2, - 35)

Maximum (- \frac{3}{2}, \frac{203}{4}), Minimum (2, - 35)

(x - 10)^{3} + 5

x^{2} + 2

\frac{1}{x - 2}

y = \frac{10 + x}{1 - x}

What is the extreme f(x)=4x^3-3x^2-36x+17 ?
The extreme f(x)=4x^3-3x^2-36x+17 is Maximum(-3/2 , 203/4),Minimum(2,-35)