What is the extreme f(x)=(x^3)/3+x^2-3x+4 ?

The extreme f(x)=(x^3)/3+x^2-3x+4 is Maximum(-3,13),Minimum(1, 7/3)

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extreme f(x)=(x^3)/3+x^2-3x+4

extreme points

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

Maximum (- 3, 13), Minimum (1, \frac{7}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 3

into

\frac{x ^{3}}{3} + x^{2} - 3 x + 4 : 13

Maximum (- 3, 13)

Plug

x = 1

into

\frac{x ^{3}}{3} + x^{2} - 3 x + 4 : \frac{7}{3}

Minimum (1, \frac{7}{3})

Maximum (- 3, 13), Minimum (1, \frac{7}{3})

f (x) = x^{2} - 10 x - 1, 2 \leq x \leq 6

f (x) = 3 3 x^{3} + 24

3 x (28 - x)

f (x) = x^{3} + 15 x - 9

f (x) = x^{3} - 2 x^{2} - 15 x + 52.2

What is the extreme f(x)=(x^3)/3+x^2-3x+4 ?
The extreme f(x)=(x^3)/3+x^2-3x+4 is Maximum(-3,13),Minimum(1, 7/3)