What is the extreme f(x)= 1/3 x^3-4x^2+12x-5 ?

The extreme f(x)= 1/3 x^3-4x^2+12x-5 is Maximum(2, 17/3),Minimum(6,-5)

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extreme f(x)= 1/3 x^3-4x^2+12x-5

extreme points

f (x) = \frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5

Maximum (2, \frac{17}{3}), Minimum (6, - 5)

Solution steps

f^{'} (x) = x^{2} - 8 x + 12

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

\frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5 : \frac{17}{3}

Maximum (2, \frac{17}{3})

Plug

x = 6

into

\frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5 : - 5

Minimum (6, - 5)

Maximum (2, \frac{17}{3}), Minimum (6, - 5)

What is the extreme f(x)= 1/3 x^3-4x^2+12x-5 ?
The extreme f(x)= 1/3 x^3-4x^2+12x-5 is Maximum(2, 17/3),Minimum(6,-5)