monotone f(x)=5x^2-(12)/(x+3)

Lösung

monotone intervalle

f (x) = 5 x^{2} - \frac{12}{x + 3}

Absteigend : - \infty < x < - 3.57903 \dots, Ansteigend : - 3.57903 \dots < x < - 3, Ansteigend : - 3 < x < - 2.27348 \dots, Absteigend : - 2.27348 \dots < x < - 0.14747 \dots, Ansteigend : - 0.14747 \dots < x < \infty

Schritte zur Lösung

Bestimme den kritischen Punkt:

x = - 3.57903 \dots, x \approx - 2.27348 \dots, x \approx - 0.14747 \dots

Bereich von

5 x^{2} - \frac{12}{x + 3} : x < - 3 or x > - 3

Intervalle finden:

Absteigend

: - \infty < x < - 3.57903 \dots,

Ansteigend

: - 3.57903 \dots < x < - 3,

Ansteigend

: - 3 < x < - 2.27348 \dots,

Absteigend

: - 2.27348 \dots < x < - 0.14747 \dots,

Ansteigend

: - 0.14747 \dots < x < \infty

Absteigend : - \infty < x < - 3.57903 \dots, Ansteigend : - 3.57903 \dots < x < - 3, Ansteigend : - 3 < x < - 2.27348 \dots, Absteigend : - 2.27348 \dots < x < - 0.14747 \dots, Ansteigend : - 0.14747 \dots < x < \infty

am pl i t u d e - sin (3 x)

d o main \frac{x}{x - 3}

d o main f (x) = \frac{x + 2}{x + 3}

in v erse f (x) = \frac{4 x + 2}{x - 9}

cr i t i c a l f (x) = \frac{x ^{2} + 17}{x - 2}