## Calculus: Early Transcendentals 8th Edition

Yes, $m_{1}m_{2}$
$f(g(x))=f(m_{2}x+b_{2})=m_{1}(m_{2}x+b_{2})+b_{1}=m_{1}m_{2}x+m_{1}b_{2}+b_{1}$ $f(g(x))$ is still a linear function with gradient $m_{1}m_{2}$