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Draw a graph of the function

$$f(x)=1-\frac{2^x}{3^x-2^x}+\left(\frac{2^x}{3^x-2^x}\right)^2...$$

and show that the function takes only the values greater than $\frac{1}{2}$.

I know that the function $f(x)$ can be expressed as an infinte sum,

$$f(x)=\sum_{i=0}^{+\infty}(-1)^i\left(\frac{2^x}{3^x-2^x}\right)^i$$

but I do not know how to deal with it.

I would be grateful for any help.

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1 Answer 1

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$f(x)$ is a geometric series. Therefore $f(x)$ equals:

$$\frac{1}{1 - -\frac{2^x}{3^x-2^x}}$$

For the geometric series to be defined, we must have $\left|-\frac{2^x}{3^x-2^x} \right| <1$. Find the $x$ that satisfy this condition, and note the shape of the graph using calculus. The conclusion follows.

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