which function increases at a faster rate on 0 to infinity, f(x) = x^2 or g(x) = 2^x? explain your reasoning.

Explanation:

Step1: Consider the growth - rate concept

Exponential functions and polynomial functions have different growth - rates as (x\to\infty). The function (f(x)=x^{2}) is a polynomial function of degree 2 and (g(x) = 2^{x}) is an exponential function.

Step2: Analyze the limit

We can use the limit (\lim_{x\to\infty}\frac{2^{x}}{x^{2}}). By applying L'Hopital's rule twice. First, if we consider the limit (\lim_{x\to\infty}\frac{2^{x}}{x^{2}}), which is in the (\frac{\infty}{\infty}) form. The derivative of (2^{x}) is (2^{x}\ln(2)) and the derivative of (x^{2}) is (2x), so (\lim_{x\to\infty}\frac{2^{x}}{x^{2}}=\lim_{x\to\infty}\frac{2^{x}\ln(2)}{2x}). This is still in the (\frac{\infty}{\infty}) form. Applying L'Hopital's rule again, the derivative of (2^{x}\ln(2)) is (2^{x}(\ln(2))^{2}) and the derivative of (2x) is 2. So (\lim_{x\to\infty}\frac{2^{x}\ln(2)}{2x}=\lim_{x\to\infty}\frac{2^{x}(\ln(2))^{2}}{2}=\infty).

Step3: Interpret the limit result

Since (\lim_{x\to\infty}\frac{2^{x}}{x^{2}}=\infty), as (x) approaches infinity, (2^{x}) grows faster than (x^{2}).

Answer:

The function (g(x)=2^{x}) increases at a faster rate on the interval ((0,\infty)) because the limit (\lim_{x\to\infty}\frac{2^{x}}{x^{2}}=\infty), indicating that (2^{x}) out - grows (x^{2}) as (x) approaches infinity.