What is the extreme f(x)= 1/3 x^3-3x^2+5x ?

The extreme f(x)= 1/3 x^3-3x^2+5x is Maximum(1, 7/3),Minimum(5,-25/3)

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extreme f(x)= 1/3 x^3-3x^2+5x

extreme points

f (x) = \frac{1}{3} x^{3} - 3 x^{2} + 5 x

Maximum (1, \frac{7}{3}), Minimum (5, - \frac{25}{3})

Solution steps

f^{'} (x) = x^{2} - 6 x + 5

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 1

into

\frac{1}{3} x^{3} - 3 x^{2} + 5 x : \frac{7}{3}

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

Plug

x = 5

into

\frac{1}{3} x^{3} - 3 x^{2} + 5 x : - \frac{25}{3}

Minimum (5, - \frac{25}{3})

Maximum (1, \frac{7}{3}), Minimum (5, - \frac{25}{3})

f (u, v) = 3 u - 7 v

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

sin (3 X)

f (x) = 2 x^{3} + 3 x^{2} - 72 x + 4

f (x) = x + cos (2 x), 0 \leq x \leq π

What is the extreme f(x)= 1/3 x^3-3x^2+5x ?
The extreme f(x)= 1/3 x^3-3x^2+5x is Maximum(1, 7/3),Minimum(5,-25/3)