What is the extreme 2/3 x^3-39/2 x^2+19x,[0<= x<= 1] ?

The extreme 2/3 x^3-39/2 x^2+19x,[0<= x<= 1] is Minimum(0,0),Maximum(1/2 , 113/24),Minimum(1, 1/6)

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extreme 2/3 x^3-39/2 x^2+19x,[0<= x<= 1]

extreme points

\frac{2}{3} x^{3} - \frac{39}{2} x^{2} + 19 x, [0 \leq x \leq 1]

Minimum (0, 0), Maximum (\frac{1}{2}, \frac{113}{24}), Minimum (1, \frac{1}{6})

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}, x = 1

Domain of

\frac{2}{3} x^{3} - \frac{39}{2} x^{2} + 19 x, 0 \leq x \leq 1 : 0 \leq x \leq 1

Find intervals:

Increasing

: 0 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < 1

Plug the extreme point

x = 0

into

\frac{2}{3} x^{3} - \frac{39}{2} x^{2} + 19 x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{2}

into

\frac{2}{3} x^{3} - \frac{39}{2} x^{2} + 19 x \Rightarrow y = \frac{113}{24}

Maximum (\frac{1}{2}, \frac{113}{24})

Plug the extreme point

x = 1

into

\frac{2}{3} x^{3} - \frac{39}{2} x^{2} + 19 x \Rightarrow y = \frac{1}{6}

Minimum (1, \frac{1}{6})

Minimum (0, 0), Maximum (\frac{1}{2}, \frac{113}{24}), Minimum (1, \frac{1}{6})

\frac{3 + x}{9 - x}

f (x) = 3 x^{3} e^{- x}, - 1 \leq x \leq 4

3 x^{4} + 12 x^{3}

f (x) = 9 x^{\frac{2}{3}} - x

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

What is the extreme 2/3 x^3-39/2 x^2+19x,[0<= x<= 1] ?
The extreme 2/3 x^3-39/2 x^2+19x,[0<= x<= 1] is Minimum(0,0),Maximum(1/2 , 113/24),Minimum(1, 1/6)