Answer:
g(x) rises more steeply than f(x)
Step-by-step explanation:
Given: [tex]f(x)=\sqrt{16x}[/tex]
[tex]g(x)=16\sqrt{x}[/tex]
The parent function of both f(x) and g(x) is same.
[tex]\text{Parent function }:p(x)=\sqrt{x}[/tex]
[tex]f(x)=\sqrt{16x}=4\sqrt{x}[/tex]
[tex]g(x)=16\sqrt{x}[/tex]
[tex]y=a\cdot f(x)[/tex]
The rise of function depends on the value of a.
Higher the value of a rises more because coefficient is more.
[tex]p(x)=\sqrt{x}[/tex]
[tex]f(x)=4\sqrt{x}[/tex]
[tex]g(x)=16\sqrt{x}[/tex]
The coefficient of g(x) is more than f(x).
We can see graphically also. Please see attachment for graph.
Hence, g(x) rises more steeply than f(x).
