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})^{…

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})^{-x}\n f(x)=\frac{3}{2}(\frac{8}{7})^{-x}\n f(x)=\frac{1}{3}(-\frac{9}{2})^{x}

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})^{-x}\n f(x)=\frac{3}{2}(\frac{8}{7})^{-x}\n f(x)=\frac{1}{3}(-\frac{9}{2})^{x}

Answer

Explanation:

Step1: Recall exponential - decay form

The general form of an exponential - decay function is $y = a\cdot b^{x}$, where $a>0$ and $0 < b<1$. We can also rewrite functions of the form $y = a\cdot(\frac{1}{b})^{-x}$ to the standard form.

Step2: Analyze each option

  • For $f(x)=\frac{3}{4}(\frac{7}{4})^{x}$, since $\frac{7}{4}>1$, it is an exponential - growth function.
  • For $f(x)=\frac{2}{3}(\frac{4}{5})^{-x}=\frac{2}{3}(\frac{5}{4})^{x}$, since $\frac{5}{4}>1$, it is an exponential - growth function.
  • For $f(x)=\frac{3}{2}(\frac{8}{7})^{-x}=\frac{3}{2}(\frac{7}{8})^{x}$. Here, $a = \frac{3}{2}>0$ and $b=\frac{7}{8}$, and $0<\frac{7}{8}<1$. So it is an exponential - decay function.
  • For $f(x)=\frac{1}{3}(-\frac{9}{2})^{x}$, the base $-\frac{9}{2}<0$, and exponential functions with a negative base are not well - defined for all real numbers $x$.

Answer:

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