which is an exponential decay function?\n$f(x)=\frac{3}{4}(\frac{7}{4})^x$\n$f(x)=\frac{2}{3}(\frac{4}{5})^{…

Explanation:

Step1: Recall exponential - decay form

The general form of an exponential - decay function is $y = a\cdot b^{x}$, where $a>0$ and $0 < b<1$, or $y=a\cdot\left(\frac{1}{b}\right)^{-x}$ where $a > 0$ and $b>1$.

Step2: Analyze option 1

For $f(x)=\frac{3}{4}\left(\frac{7}{4}\right)^{x}$, since $\frac{7}{4}>1$, it is an exponential - growth function.

Step3: Analyze option 2

For $f(x)=\frac{2}{3}\left(\frac{4}{5}\right)^{-x}=\frac{2}{3}\left(\frac{5}{4}\right)^{x}$, since $\frac{5}{4}>1$, it is an exponential - growth function.

Step4: Analyze option 3

For $f(x)=\frac{3}{2}\left(\frac{8}{7}\right)^{-x}=\frac{3}{2}\left(\frac{7}{8}\right)^{x}$. Here, $a = \frac{3}{2}>0$ and $b=\frac{7}{8}$, and $0<\frac{7}{8}<1$. So it is an exponential - decay function.

Step5: Analyze option 4

For $f(x)=\frac{1}{3}\left(-\frac{9}{2}\right)^{x}$, the base $-\frac{9}{2}<0$, and exponential functions with negative bases are not well - defined for all real numbers $x$.

Answer:

$f(x)=\frac{3}{2}\left(\frac{8}{7}\right)^{-x}$