振幅 1/9 sin(7x+pi/2)

解答

振幅

\frac{1}{9} sin (7 x + \frac{π}{2})

0.11111 \dots

求解步骤

f (x) = \frac{1}{9} sin (7 x + \frac{π}{2}), A = \frac{1}{9}

因此，振幅 |A| 为

0.11111 \dots

0.11111 \dots

in v erse \frac{2 x + 3}{5 x + 4}

in t erce pt s y = x - 4

a sy m pt o t es f (x) = \frac{x ^{2} + x - 9}{x - 2}

d o main f (x) = \frac{3 x + 5}{2 x - 3}

cr i t i c a l x^{4} - 12 x^{3} + 48 x^{2} - 64 x