What is the extreme f(x)=(x^3)/3+(x^2)/2-2x+8 ?

The extreme f(x)=(x^3)/3+(x^2)/2-2x+8 is Maximum(-2, 34/3),Minimum(1, 41/6)

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extreme f(x)=(x^3)/3+(x^2)/2-2x+8

extreme points

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

Maximum (- 2, \frac{34}{3}), Minimum (1, \frac{41}{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 + 8 : \frac{34}{3}

Maximum (- 2, \frac{34}{3})

Plug

x = 1

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 8 : \frac{41}{6}

Minimum (1, \frac{41}{6})

Maximum (- 2, \frac{34}{3}), Minimum (1, \frac{41}{6})

f (x) = - 3 x^{2} + 7 x - y^{2} + 4 y + 7

f (x) = 20 e^{- 0.1 \cdot 10}

f (x) = ∣ x y ∣

f (x) = - 2 x^{2} + 12 - 25

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

What is the extreme f(x)=(x^3)/3+(x^2)/2-2x+8 ?
The extreme f(x)=(x^3)/3+(x^2)/2-2x+8 is Maximum(-2, 34/3),Minimum(1, 41/6)