which function represents exponential decay?\n f(x)=\frac{1}{2}(\frac{3}{2})^x \n f(x)=\frac{1}{2}(-\frac{3}{…

which function represents exponential decay?\n f(x)=\frac{1}{2}(\frac{3}{2})^x \n f(x)=\frac{1}{2}(-\frac{3}{2})^x \n f(x)=4(-\frac{2}{3})^x \n f(x)=4(\frac{2}{3})^x

which function represents exponential decay?\n f(x)=\frac{1}{2}(\frac{3}{2})^x \n f(x)=\frac{1}{2}(-\frac{3}{2})^x \n f(x)=4(-\frac{2}{3})^x \n f(x)=4(\frac{2}{3})^x

Answer

Answer:

$f(x) = 4(\frac{2}{3})^x$

Explanation:

Step1: Recall exponential - decay formula

The general form of an exponential function is $y = a\cdot b^x$, where $a\neq0$, $b > 0$, and $b\neq1$. For exponential decay, $0 < b<1$.

Step2: Analyze each option

  • For $f(x)=\frac{1}{2}(\frac{3}{2})^x$, since $b = \frac{3}{2}>1$, it represents exponential growth.
  • For $f(x)=\frac{1}{2}(-\frac{3}{2})^x$ and $f(x)=4(-\frac{2}{3})^x$, the base $b$ is negative, and these are not exponential functions in the standard sense (exponential functions have positive bases).
  • For $f(x)=4(\frac{2}{3})^x$, here $a = 4$ and $b=\frac{2}{3}$, and since $0<\frac{2}{3}<1$, it represents exponential decay.