extreme (2x^2-14x+24)/(x^2+6x+40)

Lösung

extrempunkte

\frac{2 x ^{2} - 14 x + 24}{x ^{2} + 6 x + 40}

Maximum (- \frac{4 ( 7 + 335 )}{13}, \frac{73 + 4 335}{31}), Minimum (\frac{4 ( 335 - 7 )}{13}, \frac{73 - 4 335}{31})

Schritte zur Lösung

f^{'} (x) = \frac{26 x ^{2} + 112 x - 704}{( x ^{2} + 6 x + 40 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{4 ( 7 + 335 )}{13},

Absteigend

: - \frac{4 ( 7 + 335 )}{13} < x < \frac{4 ( 335 - 7 )}{13},

Ansteigend

: \frac{4 ( 335 - 7 )}{13} < x < \infty

Stecke

x = - \frac{4 ( 7 + 335 )}{13}

\frac{2 x ^{2} - 14 x + 24}{x ^{2} + 6 x + 40} : \frac{73 + 4 335}{31}

Maximum (- \frac{4 ( 7 + 335 )}{13}, \frac{73 + 4 335}{31})

Stecke

x = \frac{4 ( 335 - 7 )}{13}

\frac{2 x ^{2} - 14 x + 24}{x ^{2} + 6 x + 40} : \frac{73 - 4 335}{31}

Minimum (\frac{4 ( 335 - 7 )}{13}, \frac{73 - 4 335}{31})

Maximum (- \frac{4 ( 7 + 335 )}{13}, \frac{73 + 4 335}{31}), Minimum (\frac{4 ( 335 - 7 )}{13}, \frac{73 - 4 335}{31})

e x t re m e f (x) = x (5 - x) (8 - x), 0 < x < 5

e x t re m e f (x) = \frac{( x ^{2} + 121 )}{2 x}

e x t re m e f (x) = - 7 + 12 x - 3 x^{2}

e x t re m e y = \frac{x ^{2} - 2 x + 1}{x ^{2} + 3}

e x t re m e f (x) = \frac{2 + 4 x ^{2}}{1 - 3 x}