extreme f(x)=2sin(5x-30)+3

Lösung

extrempunkte

f (x) = 2 sin (5 x - 30) + 3

Maximum (\frac{60 - 19 π}{10} + \frac{2 π}{5} n, 5), Minimum (\frac{60 - 17 π}{10} + \frac{2 π}{5} n, 1)

Schritte zur Lösung

Bestimme den kritischen Punkt:

x = \frac{60 - 19 π}{10} + \frac{2 π}{5} n, x = \frac{60 - 17 π}{10} + \frac{2 π}{5} n

Bereich von

2 sin (5 x - 30) + 3 : - \infty < x < \infty

Intervalle finden:

Ansteigend

: \frac{2 π}{5} n \leq x < \frac{60 - 19 π}{10} + \frac{2 π}{5} n,

Absteigend

: \frac{60 - 19 π}{10} + \frac{2 π}{5} n < x < \frac{60 - 17 π}{10} + \frac{2 π}{5} n,

Ansteigend

: \frac{60 - 17 π}{10} + \frac{2 π}{5} n < x < \frac{2 π}{5} + \frac{2 π}{5} n

Setz die Extremwerte

x = \frac{60 - 19 π}{10} + \frac{2 π}{5} n

2 sin (5 x - 30) + 3 \Rightarrow y = 5

ein

Maximum (\frac{60 - 19 π}{10} + \frac{2 π}{5} n, 5)

Setz die Extremwerte

x = \frac{60 - 17 π}{10} + \frac{2 π}{5} n

2 sin (5 x - 30) + 3 \Rightarrow y = 1

ein

Minimum (\frac{60 - 17 π}{10} + \frac{2 π}{5} n, 1)

Maximum (\frac{60 - 19 π}{10} + \frac{2 π}{5} n, 5), Minimum (\frac{60 - 17 π}{10} + \frac{2 π}{5} n, 1)

in v erse (x^{2} - x)^{3}

d o main e^{2 x}

d o main f (x) = \frac{x ^{2}}{x ^{2} + 1}

r an g e \frac{5 x + 1}{7}

in t erce pt s y = \frac{2}{3} x - 5