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

Explanation:

Step1: Consider the growth - rate concept

We can use the limit to compare the growth rates of the two functions as (x\to\infty). We calculate the limit (\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{x^{2}}{2^{x}}).

Step2: Apply L'Hopital's rule

Since (\lim_{x\to\infty}\frac{x^{2}}{2^{x}}) is in the (\frac{\infty}{\infty}) form, by L'Hopital's rule, (\lim_{x\to\infty}\frac{x^{2}}{2^{x}}=\lim_{x\to\infty}\frac{2x}{2^{x}\ln 2}). This is still in the (\frac{\infty}{\infty}) form.

Step3: Apply L'Hopital's rule again

Applying L'Hopital's rule to (\lim_{x\to\infty}\frac{2x}{2^{x}\ln 2}), we get (\lim_{x\to\infty}\frac{2}{2^{x}(\ln 2)^{2}}).

Step4: Evaluate the limit

As (x\to\infty), (2^{x}\to\infty), so (\lim_{x\to\infty}\frac{2}{2^{x}(\ln 2)^{2}} = 0). This means that (g(x)=2^{x}) grows faster than (f(x)=x^{2}) as (x\to\infty).

Answer:

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