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Beliebt Funktionen >

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

Lösung

extrempunkte

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

Lösung

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}}

Schritte zur Lösung

Bestimme den kritischen Punkt:

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

Bereich von

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Setz die Extremwerte

x = 0

in

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

ein

Maximum (0, 0)

Setz die Extremwerte

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

in

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}}

ein

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}}

Graph

Beliebte Beispiele

extreme f(x)=-2x^3+45x^2-300x,4<= x<= 11

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

extreme f(x,y)=(-20)/9 x^3-5x^2y+45y

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

extreme f(x)=(6-x)/2

e x t re m e f (x) = \frac{6 - x}{2}

extreme x^2-2x-1

e x t re m e x^{2} - 2 x - 1

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

e x t re m e f (x) = 5 x^{3} - 5 x^{2} - 5 x + 9