Question

Determine o domínio, a imagem, a amplitude e o período da função f(x)=-5sen(x/2) + 3, esboçando também o seu gráfico.

Answer 1

Boa noite Mario!

Solução!

Amplitude é a altura que a função alcança no eixo y.

O período da função seno é 2π.

$f(x)=-5sen\bigg( \dfrac{x}{2}\bigg)+3\\\\\\\\\\ D=\mathbb{R}\\\\\\\\ Amplitude=|-5|=5\\\\\\\\\ x=2\pi\\\\\\\ Peri\´odo= \dfrac{2 \pi }{2}$

$Periodo= \pi$

Para determinar a imagem,vou pegar um período positivo e um negativo e substituir na função,vou trocar f(x) por y.

$x= \pi \\\\\\ f(x)=-5sen\bigg( \dfrac{ x}{2}\bigg)+3\\\\\\\\\\\ f( \pi )=-5sen\bigg( \dfrac{ \pi }{2}\bigg)+3\\\\\\\\\ f( \pi )=-5sen\bigg( 90\º\bigg)+3\\\\\\\\\\ Sen90\º=1\\\\\\\\ f( \pi )=-5(1)+3\\\\\\\\ f( \pi )=-5+3\\\\\\\\ f( \pi )=-2\\\\\\\\\ \boxed{y=-2}$

$x=-\pi\\\\\\\f(x)=-5sen\bigg( \dfrac{x}{2}\bigg)+3\\\\\\\\\\\ f(- \pi )=-5sen\bigg( \dfrac{- \pi }{2}\bigg)+3\\\\\\\\\\\ f(- \pi )=-5sen(-90\º)+3\\\\\\\\\\\ Sen-90\º=-1\\\\\\ f(- \pi )=-5(-1)+3\\\\\\\\\\\ f(- \pi )=5+3\\\\\\\\\\\ f(- \pi )=8\\\\\ \boxed{y=8}$


$Resposta:\\\\\\ Dominio=\mathbb{R}\\\\\\\ Imagem=[-2,8]\\\\\\ Amplitude=5\\\\\\ Periodo= \pi$

Boa noite!
Bons estudos!