# How to find out the result when the inverse of a function is multiplied by the original function in the given two questions:

A problem I'm facing is understanding the concept of inverse functions when multiplied to its original function. Here are the two functions provided:

f(x) = 2x + 3

h(x) = 2x

Here are the two questions being asked based on the functions given above:

(i) $$hh^{-1}(x)$$

(ii) $$ff^{-1}(5)$$

I did part (ii) by first by first making it y = 2x + 3, and then to x = 2y + 3. After this, I made it:

$$\frac{x-3}{2}$$ = y

I put 5 in the value for x to get the y value to become 1. After this, I made it f(1), where:

(2 $$\times$$ 1) + 3 = 5

So, I finally got the answer as 5.

For part (i), I was unsure as to how I could solve it using the same method.

y = 2$$^x$$

x = 2$$^y$$

Replace x with x I think just like how I replaced the x with 5 for part (ii). This would again result in

x = 2$$^y$$

I then make it h(x), which is equal to 2x. So, my final answer is 2x. However, the answer is x. I did not exactly understand how to get this. I know that the inverse of a function when multiplied to the original gives back the same variable/number. But, I wanted to know whether I could solve part (i) in a similar way to how I solved part (ii) and use this to further solve questions related to inverse functions.

If $$f(x)=2x+3$$ then the inverse is given by $$f^{-1}(x)=\frac{1}{2}(x-3)$$ so $$f(x)\cdot f^{-1}(x)=\frac{1}{2}(x-3)(2x+3)=\frac{1}{2}(2x^2-3x-9)$$
• Is it $$f(f^{-1}(5))$$ – Dr. Sonnhard Graubner Mar 11 '19 at 15:02
• Yes, we have $$h^{-1}=\frac{1}{x}$$ so $$h(h^{-1}(x))=2\cdot \frac{1}{2}x=x$$ – Dr. Sonnhard Graubner Mar 11 '19 at 16:28