extreme C(t)= t/(4t^2+11)

Lösung

extrempunkte

C (t) = \frac{t}{4 t ^{2} + 11}

Minimum (- \frac{11}{2}, - \frac{1}{4 11}), Maximum (\frac{11}{2}, \frac{1}{4 11})

Schritte zur Lösung

f^{'} (t) = \frac{- 4 t ^{2} + 11}{( 4 t ^{2} + 11 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < t < - \frac{11}{2},

Ansteigend

: - \frac{11}{2} < t < \frac{11}{2},

Absteigend

: \frac{11}{2} < t < \infty

Stecke

t = - \frac{11}{2}

\frac{t}{4 t ^{2} + 11} : - \frac{1}{4 11}

Minimum (- \frac{11}{2}, - \frac{1}{4 11})

Stecke

t = \frac{11}{2}

\frac{t}{4 t ^{2} + 11} : \frac{1}{4 11}

Maximum (\frac{11}{2}, \frac{1}{4 11})

Minimum (- \frac{11}{2}, - \frac{1}{4 11}), Maximum (\frac{11}{2}, \frac{1}{4 11})

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