extreme f(x)=x^3-2x+1

Lösung

extrempunkte

f (x) = x^{3} - 2 x + 1

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 1), Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 1)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{2}{3},

Absteigend

: - \frac{2}{3} < x < \frac{2}{3},

Ansteigend

: \frac{2}{3} < x < \infty

Stecke

x = - \frac{2}{3}

x^{3} - 2 x + 1 : - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 1

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 1)

Stecke

x = \frac{2}{3}

x^{3} - 2 x + 1 : (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 1

Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 1)

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} + 1), Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} + 1)

d o main f (x) = \frac{1}{2} (3)^{x + 4} - 5

a sy m pt o t es f (x) = \frac{x ^{2} - 2 x - 3}{x - 2}

d o main f (x) = \frac{6 + x}{1 - 6 x}

in t erce pt s f (x) = 2 x - 5

in t erce pt s f (x) = 1 - 3 x - x^{2}