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

The extreme f(x)= 1/4 (-4x^3+2x^2+12) is Minimum(0,3),Maximum(1/3 , 163/54)

Popular >

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

extreme points

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

Minimum (0, 3), Maximum (\frac{1}{3}, \frac{163}{54})

Solution steps

f^{'} (x) = x (- 3 x + 1)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < \infty

Plug

x = 0

into

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

Minimum (0, 3)

Plug

x = \frac{1}{3}

into

\frac{1}{4} (- 4 x^{3} + 2 x^{2} + 12) : \frac{163}{54}

Maximum (\frac{1}{3}, \frac{163}{54})

Minimum (0, 3), Maximum (\frac{1}{3}, \frac{163}{54})

f (x, y) = x^{2} - 2 y^{2} + 8 x - 3 y + 1

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

f (x) = 5 sin (x) cos (x), (- π, π)

y = j + x

f (x, y) = 2 x^{2} + y^{2} - 2 x y + 5 x - 3 y + 1

What is the extreme f(x)= 1/4 (-4x^3+2x^2+12) ?
The extreme f(x)= 1/4 (-4x^3+2x^2+12) is Minimum(0,3),Maximum(1/3 , 163/54)