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

The extreme (x^3)/3+(x^2)/2-2x+7 is Maximum(-2, 31/3),Minimum(1, 35/6)

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

extreme points

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 7

Maximum (- 2, \frac{31}{3}), Minimum (1, \frac{35}{6})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 7 : \frac{31}{3}

Maximum (- 2, \frac{31}{3})

Plug

x = 1

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 7 : \frac{35}{6}

Minimum (1, \frac{35}{6})

Maximum (- 2, \frac{31}{3}), Minimum (1, \frac{35}{6})

8 x + 6 y = - 5

5 x + 4 y - 15

x^{2} + 6 x + 5

y = (x - 7)^{3}

f (x) = \frac{x + 5}{- 2 x}

What is the extreme (x^3)/3+(x^2)/2-2x+7 ?
The extreme (x^3)/3+(x^2)/2-2x+7 is Maximum(-2, 31/3),Minimum(1, 35/6)