extreme f(x)=4-x^3+5x

Lösung

extrempunkte

f (x) = 4 - x^{3} + 5 x

Minimum (- \frac{5}{3}, 4 + (\frac{5}{3})^{\frac{3}{2}} - 5 \frac{5}{3}), Maximum (\frac{5}{3}, 4 - (\frac{5}{3})^{\frac{3}{2}} + 5 \frac{5}{3})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - \frac{5}{3},

Ansteigend

: - \frac{5}{3} < x < \frac{5}{3},

Absteigend

: \frac{5}{3} < x < \infty

Stecke

x = - \frac{5}{3}

4 - x^{3} + 5 x : 4 + (\frac{5}{3})^{\frac{3}{2}} - 5 \frac{5}{3}

Minimum (- \frac{5}{3}, 4 + (\frac{5}{3})^{\frac{3}{2}} - 5 \frac{5}{3})

Stecke

x = \frac{5}{3}

4 - x^{3} + 5 x : 4 - (\frac{5}{3})^{\frac{3}{2}} + 5 \frac{5}{3}

Maximum (\frac{5}{3}, 4 - (\frac{5}{3})^{\frac{3}{2}} + 5 \frac{5}{3})

Minimum (- \frac{5}{3}, 4 + (\frac{5}{3})^{\frac{3}{2}} - 5 \frac{5}{3}), Maximum (\frac{5}{3}, 4 - (\frac{5}{3})^{\frac{3}{2}} + 5 \frac{5}{3})

e x t re m e f (x) = 2 x^{3} - 6 x + 11

e x t re m e f (x) = sin (5 x), 0 \leq x \leq \frac{2 π}{5}

e x t re m e y = 16 x^{4} + 64 x^{3}

e x t re m e 3 x^{2} - 2 x y + y^{2} - 8 y

e x t re m e f (x) = 32 e^{4 x} x - 28 e^{4 x}