Composing two of three given functions to obtain $2^{x+1} - 1$

Question

A problem I'm facing involves three given functions. I need to compose two of them to obtain $2^{x+1}-1$. The three functions I'm allowed to use are

• $f(x) = 2x-1$
• $g(x) = 1/x$
• $h(x) = 2^x$

It would be great if there was a detailed explanation as to how you got the answer too. Thanks!

Answer 1

Here's the detailed explanation as to how to get the answer:

Compute all applicable compositions, i.e., $f(g(x))$, $g(f(x))$, $f(h(x))$, $h(f(x))$, $g(h(x))$, $h(g(x))$. Then check if any of these equals (or can be simplified to) the desired function of $x$.

Answer 2

But $$h(f(x))=2^{2x-1},h(g(x))=2^{1/x}$$ which gives not your solution, so $$f(h(x))$$ leads to your solutions

Answer 3

Hint:

Consider the functions $f$ and $h$. This answer will be more complicated than simply plugging things in, you'll need to make use of some algebraic properties.

In particular, the property that $$a^b a^c = a^{b+c}$$

---

Solution:

So, $f(x) = 2x-1, h(x) = 2^x$. We know we want to get two to the something, so we should apply $h$ at some point, and we kind of want a $-1$ to appear at some point, so that's why we choose $f$.

Sadly beyond that stuff like this is largely guess and check (such as other possible functions or the order to apply them); I'll say right off that the right composition is $f \circ h$.

This gives us the function composition

$$(f\circ h)(x) = f\left( h(x) \right) = 2\cdot 2^x - 1$$

We make use of the property I noted earlier: $2\cdot 2^x = 2^{x+1}$. Thus

$$(f\circ h)(x) = 2^{x+1} - 1$$