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

The extreme (x^4)/4-(x^3)/3-2x^2+4x+3 is Minimum(-2,-19/3),Maximum(1, 59/12),Minimum(2, 13/3)

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

extreme points

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

Minimum (- 2, - \frac{19}{3}), Maximum (1, \frac{59}{12}), Minimum (2, \frac{13}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} - 2 x^{2} + 4 x + 3 : - \frac{19}{3}

Minimum (- 2, - \frac{19}{3})

Plug

x = 1

into

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} - 2 x^{2} + 4 x + 3 : \frac{59}{12}

Maximum (1, \frac{59}{12})

Plug

x = 2

into

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

Minimum (2, \frac{13}{3})

Minimum (- 2, - \frac{19}{3}), Maximum (1, \frac{59}{12}), Minimum (2, \frac{13}{3})

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

f (x, y) = 4 x y - x^{2} - y^{2} - 14 x + 4 y + 10

f (x) = x^{3} + 14 x - 3

y = \frac{sin ( x )}{x} {- 2, π \leq x \leq 2 π}

2 x^{4} + y^{2} - x^{2} - 2 y

What is the extreme (x^4)/4-(x^3)/3-2x^2+4x+3 ?
The extreme (x^4)/4-(x^3)/3-2x^2+4x+3 is Minimum(-2,-19/3),Maximum(1, 59/12),Minimum(2, 13/3)