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

Explanation:

Step1: Recall the form of exponential decay function

The general form of an exponential function is (y = a\cdot b^{x}), where (a>0). For it to be a decay function, (0 < b<1). Also, we can use the property (b^{-x}=\left(\frac{1}{b}\right)^{x}).

Step2: Analyze each function

• For (f(x)=\frac{3}{4}\left(\frac{7}{4}\right)^{x}), here (a = \frac{3}{4}>0) and (b=\frac{7}{4}>1), so it is an exponential growth function.
• For (f(x)=\frac{2}{3}\left(\frac{4}{5}\right)^{-x}), using the property (b^{-x}=\left(\frac{1}{b}\right)^{x}), we have (f(x)=\frac{2}{3}\left(\frac{5}{4}\right)^{x}), where (a=\frac{2}{3}>0) and (b = \frac{5}{4}>1), so it is an exponential growth function.
• For (f(x)=\frac{3}{2}\left(\frac{8}{7}\right)^{-x}), using the property (b^{-x}=\left(\frac{1}{b}\right)^{x}), we get (f(x)=\frac{3}{2}\left(\frac{7}{8}\right)^{x}), where (a=\frac{3}{2}>0) and (0<\frac{7}{8}<1), so it is an exponential decay function.
• For (f(x)=\frac{1}{3}\left(-\frac{9}{2}\right)^{x}), since the base (b =-\frac{9}{2}<0), this is not a valid exponential function (because for real - valued exponential functions, the base (b>0,b\neq1)).

Answer:

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