extreme f(x)=e^{(x)}(2x^2+x-8)

Lösung

extrempunkte

f (x) = e^{(x)} (2 x^{2} + x - 8)

Maximum (- \frac{7}{2}, \frac{13}{e ^{\frac{7}{2}}}), Minimum (1, - 5 e)

Schritte zur Lösung

f^{'} (x) = e^{x} (x - 1) (2 x + 7)

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{7}{2},

Absteigend

: - \frac{7}{2} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{7}{2}

e^{x} (2 x^{2} + x - 8) : \frac{13}{e ^{\frac{7}{2}}}

Maximum (- \frac{7}{2}, \frac{13}{e ^{\frac{7}{2}}})

Stecke

x = 1

e^{x} (2 x^{2} + x - 8) : - 5 e

Minimum (1, - 5 e)

Maximum (- \frac{7}{2}, \frac{13}{e ^{\frac{7}{2}}}), Minimum (1, - 5 e)

in v erse f (x) = \frac{5 - x}{x}

mi d p o in t (- 2, - 5), (3, - 2)

d o main - \frac{1}{2 9 - x}

p a r a ll e l 10 x + 6 y = 8

y = 2 x - 3