What is the extreme f(x)=((-8x^3+3x^2))/4 ?

The extreme f(x)=((-8x^3+3x^2))/4 is Minimum(0,0),Maximum(1/4 , 1/64)

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extreme f(x)=((-8x^3+3x^2))/4

extreme points

f (x) = \frac{( - 8 x ^{3} + 3 x ^{2} )}{4}

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{1}{64})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < \infty

Plug

x = 0

into

\frac{- 8 x ^{3} + 3 x ^{2}}{4} : 0

Minimum (0, 0)

Plug

x = \frac{1}{4}

into

\frac{- 8 x ^{3} + 3 x ^{2}}{4} : \frac{1}{64}

Maximum (\frac{1}{4}, \frac{1}{64})

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{1}{64})

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

f (x) = (5 + x)^{5}

f (x, y) = y (e^{x} - 1)

f (x) = \frac{x}{x ^{2} + 9}, - 4 \leq x \leq 5

x^{3} + 3 x^{2} - 24 x

What is the extreme f(x)=((-8x^3+3x^2))/4 ?
The extreme f(x)=((-8x^3+3x^2))/4 is Minimum(0,0),Maximum(1/4 , 1/64)