extreme f(x)=x^3-7x-6

Lösung

extrempunkte

f (x) = x^{3} - 7 x - 6

Maximum (- \frac{7}{3}, - (\frac{7}{3})^{\frac{3}{2}} + 7 \frac{7}{3} - 6), Minimum (\frac{7}{3}, (\frac{7}{3})^{\frac{3}{2}} - 7 \frac{7}{3} - 6)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 7

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{7}{3},

Absteigend

: - \frac{7}{3} < x < \frac{7}{3},

Ansteigend

: \frac{7}{3} < x < \infty

Stecke

x = - \frac{7}{3}

x^{3} - 7 x - 6 : - (\frac{7}{3})^{\frac{3}{2}} + 7 \frac{7}{3} - 6

Maximum (- \frac{7}{3}, - (\frac{7}{3})^{\frac{3}{2}} + 7 \frac{7}{3} - 6)

Stecke

x = \frac{7}{3}

x^{3} - 7 x - 6 : (\frac{7}{3})^{\frac{3}{2}} - 7 \frac{7}{3} - 6

Minimum (\frac{7}{3}, (\frac{7}{3})^{\frac{3}{2}} - 7 \frac{7}{3} - 6)

Maximum (- \frac{7}{3}, - (\frac{7}{3})^{\frac{3}{2}} + 7 \frac{7}{3} - 6), Minimum (\frac{7}{3}, (\frac{7}{3})^{\frac{3}{2}} - 7 \frac{7}{3} - 6)

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

e x t re m e f (x, y) = 4 x^{2} + 5 x y + 5 y^{2}

e x t re m e - 2 x^{2} y + 6 x^{2} + 4 y^{2} - 16 y + 16

e x t re m e f (x) = \frac{x ^{2} + 7 x + 3}{x + 2}

e x t re m e (5 + x)^{5}