what is the rate of increase for the function f(x) = $\frac{1}{3}(sqrt3{24})^{2x}$?\n$\frac{1}{3}$\n$2sqrt3{3…

Answer:

C. 4

Explanation:

Step1: Recall exponential - function form

The general form of an exponential function is $y = a\cdot b^{x}$, where $b$ is the growth factor (rate of increase when $b>1$). The given function is $f(x)=\frac{1}{3}(\sqrt[3]{24})^{2x}$. We can rewrite it using the exponent rule $(a^{m})^{n}=a^{mn}$.

Step2: Rewrite the function

$f(x)=\frac{1}{3}(\sqrt[3]{24})^{2x}=\frac{1}{3}(24^{\frac{1}{3}})^{2x}=\frac{1}{3}\cdot24^{\frac{2x}{3}}$. Now, $24 = 8\times3$, so $24^{\frac{2x}{3}}=(8\times3)^{\frac{2x}{3}}$. Using the rule $(ab)^{n}=a^{n}b^{n}$, we have $(8\times3)^{\frac{2x}{3}}=8^{\frac{2x}{3}}\times3^{\frac{2x}{3}}$. Since $8 = 2^{3}$, then $8^{\frac{2x}{3}}=(2^{3})^{\frac{2x}{3}}=2^{2x}=(2^{2})^{x}=4^{x}$. So $f(x)=\frac{1}{3}\times4^{x}\times3^{\frac{2x}{3}}$. The part that determines the rate of increase is the base of the exponential part in the form $y = a\cdot b^{x}$, and for $y = \frac{1}{3}\times4^{x}\times3^{\frac{2x}{3}}$, the base of the main - exponential part (the part that gives the rate of increase) is 4.