Final answer:
The function g(x) increases at a faster rate than f(x) because g(x) has a larger exponent in its exponential term and a coefficient of 3, which multiplies the whole function, leading to a steeper slope as x increases.
Step-by-step explanation:
The relationship between the graphs of the functions f(x) = \sqrt{16^x} and g(x) = 3\sqrt{64^x} can be examined by simplifying both expressions. First, recognize that 16 is 2 raised to the 4th power (2^4), and 64 is 2 raised to the 6th power (2^6). Therefore, we can rewrite the functions as follows:
- f(x) = \sqrt{(2^4)^x} = \sqrt{2^{4x}} = 2^{2x}
- g(x) = 3\sqrt{(2^6)^x} = 3\sqrt{2^{6x}} = 3 \cdot 2^{3x}
Now, by comparing the exponents, we see that f(x) is an exponential function with a base of 2 and an exponent of 2x, while g(x) is also an exponential function but with a coefficient of 3 and an exponent of 3x. The initial value (the value when x = 0) of both functions is 1 and 3, respectively, because 20 = 1 and 3 \cdot 20 = 3. The presence of the coefficient 3 in g(x) means that g(x) will increase at a faster rate compared to f(x) as x increases. Therefore, the correct relationship between f(x) and g(x) is that g(x) increases at a faster rate than f(x) due to the higher exponent in the exponential term and the multiplicative coefficient 3.