monotone f(x)= 1/(x^2-9)

Lösung

monotone intervalle

f (x) = \frac{1}{x ^{2} - 9}

Ansteigend : - \infty < x < - 3, Ansteigend : - 3 < x < 0, Absteigend : 0 < x < 3, Absteigend : 3 < x < \infty

Schritte zur Lösung

f^{'} (x) = - \frac{2 x}{( x ^{2} - 9 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Ansteigend

: - 3 < x < 0,

Absteigend

: 0 < x < 3,

Absteigend

: 3 < x < \infty

Ansteigend : - \infty < x < - 3, Ansteigend : - 3 < x < 0, Absteigend : 0 < x < 3, Absteigend : 3 < x < \infty

p a r i t y f (x) = 7 x^{4} - 2 x^{3}

d o main \frac{3 x ^{2} - 9 x + 12}{x ^{2} - 10 x + 25}

p er i o d i c i t y y = 3 cot (\frac{1}{2} x) - 2

d o main \frac{1}{1 + 5 ( \frac{1 - x}{5 x} )}

r an g e X^{3}