What is the extreme 1/2 (x^3-3x+7) ?

The extreme 1/2 (x^3-3x+7) is Maximum(-1, 9/2),Minimum(1, 5/2)

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extreme 1/2 (x^3-3x+7)

extreme points

\frac{1}{2} (x^{3} - 3 x + 7)

Maximum (- 1, \frac{9}{2}), Minimum (1, \frac{5}{2})

Solution steps

f^{'} (x) = \frac{1}{2} (3 x^{2} - 3)

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{1}{2} (x^{3} - 3 x + 7) : \frac{9}{2}

Maximum (- 1, \frac{9}{2})

Plug

x = 1

into

\frac{1}{2} (x^{3} - 3 x + 7) : \frac{5}{2}

Minimum (1, \frac{5}{2})

Maximum (- 1, \frac{9}{2}), Minimum (1, \frac{5}{2})

f (x) = \frac{e ^{x}}{x ^{3}}

f (x) = \frac{( x ^{2} - 8 )}{x - 3}

q (x, y) = x^{2} - x y + 5 y^{2}

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

f (x) = 5 x^{3} - 4

What is the extreme 1/2 (x^3-3x+7) ?
The extreme 1/2 (x^3-3x+7) is Maximum(-1, 9/2),Minimum(1, 5/2)