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

Lösung

extrempunkte

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

Maximum (- \frac{1 + 10}{3}, \frac{1 + 10}{2}), Minimum (\frac{- 1 + 10}{3}, \frac{1 - 10}{2})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{1 + 10}{3},

Absteigend

: - \frac{1 + 10}{3} < x < \frac{10 - 1}{3},

Ansteigend

: \frac{10 - 1}{3} < x < \infty

Stecke

x = - \frac{1 + 10}{3}

\frac{x ^{2} - 3 x}{x ^{2} + 1} : \frac{1 + 10}{2}

Maximum (- \frac{1 + 10}{3}, \frac{1 + 10}{2})

Stecke

x = \frac{- 1 + 10}{3}

\frac{x ^{2} - 3 x}{x ^{2} + 1} : \frac{1 - 10}{2}

Minimum (\frac{- 1 + 10}{3}, \frac{1 - 10}{2})

Maximum (- \frac{1 + 10}{3}, \frac{1 + 10}{2}), Minimum (\frac{- 1 + 10}{3}, \frac{1 - 10}{2})

e x t re m e 4 x^{5} - 5 x^{4}

e x t re m e f (x) = 4 x^{2} + 8 x + 4 y^{2} + 4

e x t re m e f (x) = 2 x + 1

e x t re m e f (x, y) = 2 x + 3 y + 5

e x t re m e f (x) = x^{4} + 4 x^{3} + 5