What is the extreme f(x)= 1/3 x^3-x^2-3x+4 ?

The extreme f(x)= 1/3 x^3-x^2-3x+4 is Maximum(-1, 17/3),Minimum(3,-5)

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extreme f(x)= 1/3 x^3-x^2-3x+4

extreme points

f (x) = \frac{1}{3} x^{3} - x^{2} - 3 x + 4

Maximum (- 1, \frac{17}{3}), Minimum (3, - 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

\frac{1}{3} x^{3} - x^{2} - 3 x + 4 : \frac{17}{3}

Maximum (- 1, \frac{17}{3})

Plug

x = 3

into

\frac{1}{3} x^{3} - x^{2} - 3 x + 4 : - 5

Minimum (3, - 5)

Maximum (- 1, \frac{17}{3}), Minimum (3, - 5)

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

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

f (x) = - (x + 4)^{3}, - 4 \leq x \leq - 3

f (x) = \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} + 10 x - 4

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

What is the extreme f(x)= 1/3 x^3-x^2-3x+4 ?
The extreme f(x)= 1/3 x^3-x^2-3x+4 is Maximum(-1, 17/3),Minimum(3,-5)