What is the extreme f(x)=x^4-5x^{3/2} ?

The extreme f(x)=x^4-5x^{3/2} is

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extreme f(x)=x^4-5x^{3/2}

extreme points

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

Maximum (0, 0), Minimum \frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}}, \frac{22 5 ^{\frac{4}{5}}}{16 \cdot 2 ^{\frac{4}{5}}} - 5 (\frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}})^{\frac{3}{2}}

Solution steps

Find the critical points:

x = 0, x = \frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}}

Domain of

x^{4} - 5 x^{\frac{3}{2}} : x \geq 0

Find intervals:

Decreasing

: 0 < x < \frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}},

Increasing

: \frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}} < x < \infty

Plug the extreme point

x = 0

into

x^{4} - 5 x^{\frac{3}{2}} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = \frac{22 5 ^{\frac{1}{5}}}{2 2 ^{\frac{1}{5}}}

into

x^{4} - 5 x^{\frac{3}{2}} \Rightarrow y = \frac{22 5 ^{\frac{4}{5}}}{16 2 ^{\frac{4}{5}}} - 5 (\frac{22 5 ^{\frac{1}{5}}}{2 2 ^{\frac{1}{5}}})^{\frac{3}{2}}

Minimum \frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}}, \frac{22 5 ^{\frac{4}{5}}}{16 \cdot 2 ^{\frac{4}{5}}} - 5 (\frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}})^{\frac{3}{2}}

Maximum (0, 0), Minimum \frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}}, \frac{22 5 ^{\frac{4}{5}}}{16 \cdot 2 ^{\frac{4}{5}}} - 5 (\frac{22 5 ^{\frac{1}{5}}}{2 \cdot 2 ^{\frac{1}{5}}})^{\frac{3}{2}}

f (x) = - 2 x^{3} + 45 x^{2} - 300 x, 4 \leq x \leq 11

f (x, y) = \frac{- 20}{9} x^{3} - 5 x^{2} y + 45 y

f (x) = \frac{6 - x}{2}

x^{2} - 2 x - 1

f (x) = 5 x^{3} - 5 x^{2} - 5 x + 9