extreme f(x)=(3000-10x^2+1)/(3x^3)

Lösung

extrempunkte

f (x) = \frac{3000 - 10 x ^{2} + 1}{3 x ^{3}}

Maximum (- \frac{9003}{10}, \frac{20 10}{9 9003}), Minimum (\frac{9003}{10}, - \frac{20 10}{9 9003})

Schritte zur Lösung

f^{'} (x) = - \frac{- 10 x ^{2} + 9003}{3 x ^{4}}

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{9003}{10},

Absteigend

: - \frac{9003}{10} < x < 0,

Absteigend

: 0 < x < \frac{9003}{10},

Ansteigend

: \frac{9003}{10} < x < \infty

Stecke

x = - \frac{9003}{10}

\frac{3000 - 10 x ^{2} + 1}{3 x ^{3}} : \frac{20 10}{9 9003}

Maximum (- \frac{9003}{10}, \frac{20 10}{9 9003})

Stecke

x = \frac{9003}{10}

\frac{3000 - 10 x ^{2} + 1}{3 x ^{3}} : - \frac{20 10}{9 9003}

Minimum (\frac{9003}{10}, - \frac{20 10}{9 9003})

Maximum (- \frac{9003}{10}, \frac{20 10}{9 9003}), Minimum (\frac{9003}{10}, - \frac{20 10}{9 9003})

in v erse f (x) = \frac{8}{5 x + 7}

s im pl i f y (1.5) (1.7)

d o main f (x) = x - 7

r an g e - 2 (x + 3)^{2} - 1

y = 2 x - 7