What is the extreme f(x)=2x^3-3x^2-12x+5 ?

The extreme f(x)=2x^3-3x^2-12x+5 is Maximum(-1,12),Minimum(2,-15)

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extreme f(x)=2x^3-3x^2-12x+5

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 5

Maximum (- 1, 12), Minimum (2, - 15)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

2 x^{3} - 3 x^{2} - 12 x + 5 : 12

Maximum (- 1, 12)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 5 : - 15

Minimum (2, - 15)

Maximum (- 1, 12), Minimum (2, - 15)

\frac{x ^{3} - 8}{x ^{2}}

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

3 e^{- 2 x}

x 36 - x^{2}

What is the extreme f(x)=2x^3-3x^2-12x+5 ?
The extreme f(x)=2x^3-3x^2-12x+5 is Maximum(-1,12),Minimum(2,-15)