\#. 5 increasing with the range $$ (2, \infty) $$ ? a. $$ f(x)=\frac{1}{2}(5)^{x}+2 $$ f. $$ f(x)=-\frac{1}{2}(5)^{x}+2 $$ c. $$ f(x)=\frac{1}{2}\left(\frac{1}{5}\right)^{x}+2 $$ d. $$ f(x)=-\frac{1}{2}\left(\frac{1}{5}\right)^{x}+2 $$

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1. For the function to be increasing, the exponential component must be increasing. An exponential function $$ a^x $$ is increasing if and only if $$ a > 1 $$.

2. Examining option a:
   $$
   f(x)=\frac{1}{2}(5)^x+2
   $$
   Here, the base is $$ 5 $$ (which is greater than 1), so $$ (5)^x $$ is increasing. Multiplying by a positive factor $$ \frac{1}{2} $$ preserves this behavior.

3. Next, we determine the range. Since exponential functions have a horizontal asymptote, we examine:
   - As $$ x\to -\infty $$, $$ (5)^x\to 0 $$, hence:
     $$
     f(x)\to \frac{1}{2}(0)+2=2.
     $$
   - As $$ x\to \infty $$, $$ (5)^x\to \infty $$, hence:
     $$
     f(x)\to \infty.
     $$
   Therefore, the range of the function is $$ (2, \infty) $$.

4. Options f, c, and d either have a negative coefficient or a base less than 1 (or both), which cause the function to be decreasing or have a range that does not match $$ (2, \infty) $$.

Thus, the correct answer is

a. $$ f(x)=\frac{1}{2}(5)^x+2 $$.
The correct function is $$ f(x)=\frac{1}{2}(5)^x+2 $$.

#. 5
increasing with the range ?
a.
f.
c.
d.

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#. 5 increasing with the range ? a. f. c. d.