both of these functions grow as x gets larger and larger. which function eventually exceeds the…

Explanation:

Step1: Analyze the growth rate of linear function

A linear function (y = mx + b) ((f(x)=\frac{15}{4}x + 3) where (m=\frac{15}{4}), (b = 3)) has a constant rate of change. Its growth rate is determined by the slope (m). As (x\to+\infty), (f(x)) grows linearly.

Step2: Analyze the growth rate of exponential function

An exponential function (y = a\cdot b^{x}) ((g(x)=\frac{3}{4}(5)^{x}) where (a=\frac{3}{4}), (b = 5>1)). The general form of an exponential function (y = a\cdot b^{x}) with (b>1) has a growth rate that is proportional to its current value. As (x) increases, the growth rate of (y=a\cdot b^{x}) gets larger and larger. Mathematically, we can use the limit (\lim_{x\rightarrow+\infty}\frac{f(x)}{g(x)}=\lim_{x\rightarrow+\infty}\frac{\frac{15}{4}x + 3}{\frac{3}{4}(5)^{x}}). We can apply L'Hopital's rule. For the limit (\lim_{x\rightarrow+\infty}\frac{\frac{15}{4}x + 3}{\frac{3}{4}(5)^{x}}), since it is in the (\frac{\infty}{\infty}) form. Differentiate the numerator and the denominator. The derivative of the numerator (u=\frac{15}{4}x + 3) is (u^\prime=\frac{15}{4}), and the derivative of the denominator (v=\frac{3}{4}(5)^{x}) is (v^\prime=\frac{3}{4}\ln(5)\cdot5^{x}). Then (\lim_{x\rightarrow+\infty}\frac{\frac{15}{4}}{\frac{3}{4}\ln(5)\cdot5^{x}} = 0)

Answer:

The function (g(x)=\frac{3}{4}(5)^{x}) eventually exceeds (f(x)=\frac{15}{4}x + 3)