extreme f(x)=(-x-8)/(10x^2-40x+27)

Lösung

extrempunkte

f (x) = \frac{- x - 8}{10 x ^{2} - 40 x + 27}

Maximum (- \frac{80 + 9870}{10}, \frac{100 - 9870}{260}), Minimum (\frac{- 80 + 9870}{10}, \frac{100 + 9870}{260})

Schritte zur Lösung

f^{'} (x) = \frac{10 x ^{2} + 160 x - 347}{( 10 x ^{2} - 40 x + 27 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{80 + 9870}{10},

Absteigend

: - \frac{80 + 9870}{10} < x < \frac{20 - 130}{10},

Absteigend

: \frac{20 - 130}{10} < x < \frac{9870 - 80}{10},

Ansteigend

: \frac{9870 - 80}{10} < x < \frac{20 + 130}{10},

Ansteigend

: \frac{20 + 130}{10} < x < \infty

Stecke

x = - \frac{80 + 9870}{10}

\frac{- x - 8}{10 x ^{2} - 40 x + 27} : \frac{100 - 9870}{260}

Maximum (- \frac{80 + 9870}{10}, \frac{100 - 9870}{260})

Stecke

x = \frac{- 80 + 9870}{10}

\frac{- x - 8}{10 x ^{2} - 40 x + 27} : \frac{100 + 9870}{260}

Minimum (\frac{- 80 + 9870}{10}, \frac{100 + 9870}{260})

Maximum (- \frac{80 + 9870}{10}, \frac{100 - 9870}{260}), Minimum (\frac{- 80 + 9870}{10}, \frac{100 + 9870}{260})

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