What is the extreme f(x)=4x^3-3x^2-18x+17 ?

The extreme f(x)=4x^3-3x^2-18x+17 is Maximum(-1,28),Minimum(3/2 ,-13/4)

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extreme f(x)=4x^3-3x^2-18x+17

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 18 x + 17

Maximum (- 1, 28), Minimum (\frac{3}{2}, - \frac{13}{4})

Solution steps

f^{'} (x) = 12 x^{2} - 6 x - 18

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

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

Increasing

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

Plug

x = - 1

into

4 x^{3} - 3 x^{2} - 18 x + 17 : 28

Maximum (- 1, 28)

Plug

x = \frac{3}{2}

into

4 x^{3} - 3 x^{2} - 18 x + 17 : - \frac{13}{4}

Minimum (\frac{3}{2}, - \frac{13}{4})

Maximum (- 1, 28), Minimum (\frac{3}{2}, - \frac{13}{4})

(4 - x)^{2} + 6

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

x + 2 - 2

(0, 2), (8, 8)

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

What is the extreme f(x)=4x^3-3x^2-18x+17 ?
The extreme f(x)=4x^3-3x^2-18x+17 is Maximum(-1,28),Minimum(3/2 ,-13/4)