Answer: $$ f(x)=-2\left(\frac{1}{5}\right)+2 $$ functions is decreasing? a.. $$ f(x)=\frac{1}{2}(3)^{x}-8 $$ b. $$ f(x)=-\frac{1}{2}\left(\frac{1}{3}\right)^{x}-8 $$ c. $$ f(x)=-\frac{1}{2}(3)^{x}-8 $$ d. $$ f(x)=2(3)^{x}-8 $$

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We are given a list of exponential functions and asked which one is decreasing. An exponential function of the form $$ f(x)=A\cdot b^x+C $$ has derivative $$ f'(x)=A\cdot b^x\ln(b). $$ Since $$ b^x>0 $$ for all $$ x $$, the sign of $$ f'(x) $$ depends on the product $$ A\ln(b) $$. A function is decreasing when $$ f'(x)<0 $$ for all $$ x $$. This requires: 1. $$ A\ln(b)<0 $$. Let’s analyze each option: 1. **Option a**: $$ f(x)=\frac{1}{2}(3)^x-8. $$ Here, $$ A=\frac{1}{2}>0 $$ and $$ b=3>1 $$ so that $$\ln(3)>0$$. Thus, $$ A\ln(b)=\frac{1}{2}\ln(3)>0, $$ so the function is increasing. 2. **Option b**: $$ f(x)=-\frac{1}{2}\left(\frac{1}{3}\right)^x-8. $$ Here, $$ A=-\frac{1}{2}<0 $$ and $$ b=\frac{1}{3}<1 $$, so $$\ln\left(\frac{1}{3}\right)<0$$. Thus, $$ A\ln(b)=-\frac{1}{2}\cdot\ln\left(\frac{1}{3}\right). $$ Since the product of two negatives is positive, $$ A\ln(b)>0 $$, and the function is increasing. 3. **Option c**: $$ f(x)=-\frac{1}{2}(3)^x-8. $$ Here, $$ A=-\frac{1}{2}<0 $$ and $$ b=3>1 $$ so that $$\ln(3)>0$$. Thus, $$ A\ln(b)=-\frac{1}{2}\ln(3)<0, $$ which means the function is decreasing. 4. **Option d**: $$ f(x)=2(3)^x-8. $$ Here, $$ A=2>0 $$ and $$ b=3>1 $$ so that $$\ln(3)>0$$. Thus, $$ A\ln(b)=2\ln(3)>0, $$ and the function is increasing. The only function that satisfies $$ A\ln(b)<0 $$ is option **c**. Additionally, the expression $$ f(x)=-2\left(\frac{1}{5}\right)+2 $$ appears to be a separate calculation where $$ -2\left(\frac{1}{5}\right)=-\frac{2}{5} $$ and $$ -\frac{2}{5}+2=\frac{8}{5}. $$ This expression does not affect the analysis of the exponential functions but is an evaluated numerical expression. So, the decreasing exponential function is: $$ \boxed{f(x)=-\frac{1}{2}(3)^x-8.} $$ The decreasing function is $$ f(x)=-\frac{1}{2}(3)^x-8 $$.

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c.
d.

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