What is the extreme y=y=(x^3)/3+(x^2)/2-2x+9 ?

The extreme y=y=(x^3)/3+(x^2)/2-2x+9 is Maximum(-2, 37/3),Minimum(1, 47/6)

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extreme y=y=(x^3)/3+(x^2)/2-2x+9

extreme points

y = y = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 9

Maximum (- 2, \frac{37}{3}), Minimum (1, \frac{47}{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 + 9 : \frac{37}{3}

Maximum (- 2, \frac{37}{3})

Plug

x = 1

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 9 : \frac{47}{6}

Minimum (1, \frac{47}{6})

Maximum (- 2, \frac{37}{3}), Minimum (1, \frac{47}{6})

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

f (x, y) = - x^{4} + 8 x^{2} - y^{2} - 1

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

\frac{x + 5}{2 - x}

f (x) = x^{2} - 8

What is the extreme y=y=(x^3)/3+(x^2)/2-2x+9 ?
The extreme y=y=(x^3)/3+(x^2)/2-2x+9 is Maximum(-2, 37/3),Minimum(1, 47/6)