What is the extreme y=2x^3-3x+11 ?

The extreme y=2x^3-3x+11 is Maximum(-1/(sqrt(2)),sqrt(2)+11),Minimum(1/(sqrt(2)),-sqrt(2)+11)

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extreme y=2x^3-3x+11

extreme points

y = 2 x^{3} - 3 x + 11

Maximum (- \frac{1}{2}, 2 + 11), Minimum (\frac{1}{2}, - 2 + 11)

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

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

Plug

x = - \frac{1}{2}

into

2 x^{3} - 3 x + 11 : 2 + 11

Maximum (- \frac{1}{2}, 2 + 11)

Plug

x = \frac{1}{2}

into

2 x^{3} - 3 x + 11 : - 2 + 11

Minimum (\frac{1}{2}, - 2 + 11)

Maximum (- \frac{1}{2}, 2 + 11), Minimum (\frac{1}{2}, - 2 + 11)

y = - x^{2} + 1, - 2 \leq x \leq 1

f (x) = x^{3} - 6 x^{2} - 15 x

f (x) = \frac{( x ^{2} - 3 )}{x - 2}

f (x) = 4 x + \frac{9}{x}

f (x) = x^{3} + 3 x^{2} + 3 x

What is the extreme y=2x^3-3x+11 ?
The extreme y=2x^3-3x+11 is Maximum(-1/(sqrt(2)),sqrt(2)+11),Minimum(1/(sqrt(2)),-sqrt(2)+11)