The general form of an exponential function is:
[tex]f(x)=ab^x[/tex]We are given the points:
[tex](-1,\frac{1}{2})\text{ }and\text{ }(1,8)[/tex]This means:
[tex]\begin{gathered} f(-1)=\frac{1}{2} \\ . \\ f(1)=8 \end{gathered}[/tex]Thus, we can write:
[tex]\begin{cases}\frac{1}{2}=ab^{-1}{} \\ 8=ab^1\end{cases}[/tex]This is a system of two equations with two variables. We can solve for b in the first equation:
[tex]\begin{gathered} \frac{1}{2}=ab^{-1} \\ . \\ \frac{1}{2}=\frac{a}{b} \\ . \\ b=2a \end{gathered}[/tex]And now, substitute in the second equation:
[tex]\begin{gathered} 8=a(2a)^1 \\ . \\ 8=2a^2 \\ . \\ \frac{8}{2}=a^2 \\ . \\ a=\pm\sqrt{4} \\ . \\ a=\pm2 \end{gathered}[/tex]Now, we can pick any of the two solutions for a, in this case, we'll use the positive one, a = 2. And now we can find b:
[tex]\begin{gathered} b=2\cdot2 \\ . \\ b=4 \end{gathered}[/tex]Now, we can write the equation:
[tex]f(x)=2\cdot4^x[/tex]And we can verify that the points given lie in the graph of f(x), by evaluating:
[tex]\begin{gathered} f(-1)=2\cdot4^{-1}=\frac{2}{4}=\frac{1}{2} \\ . \\ f(1)=2\cdot4^1=2\cdot4=8 \end{gathered}[/tex]Thus, the answer:
[tex]f(x)=2\cdot4^x[/tex]is correct.