Answer:
The function g(x) increases at a faster rate
The function g(x) has a greater initial value
Step-by-step explanation:
we have
[tex]f(x)=(\sqrt{16})^{x}[/tex]
[tex]g(x)=3(\sqrt{64})^{x}[/tex]
we know that
[tex]16=4^{2}[/tex]
substitute in f(x)
[tex]f(x)=(\sqrt{4^{2}})^{x}=4^x[/tex]
[tex]64=8^{2}[/tex]
substitute in g(x)
[tex]g(x)=3(\sqrt{8^{2}})^{x}=3(8)^x[/tex]
Compare the functions
f(x) and g(x) are not equivalent
Find the initial value
For x=0
[tex]f(0)=4^0=1[/tex]
[tex]g(0)=3(8)^0=3[/tex]
The initial value of g(x) is greater than the initial value of f(x)
using a graphing tool
The function g(x) increases at a faster rate than the function f(x) (see the attached figure)
That means ----> The value of g(x) is greater than the value of f(x) for the same value of x
therefore
The function g(x) increases at a faster rate
The function g(x) has a greater initial value
