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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.

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If $$f(x)=2x+3$$ then the inverse is given by $$f^{-1}(x)=\frac{1}{2}(x-3)$$ so $$f(x)\cdot f^{-1}(x)=\frac{1}{2}(x-3)(2x+3)=\frac{1}{2}(2x^2-3x-9)$$

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  • $\begingroup$ Thank you for your answer. I had a few doubts though. Is it possible to get the value for part (i) in the same way as how you have solved this? And to get the final answer, do we have to substitute 5 into the equation you have formed? Also, is it possible to solve part (i) in a similar way to how I solved part (ii)? $\endgroup$ – V11 Mar 11 at 14:53
  • $\begingroup$ Is it $$f(f^{-1}(5))$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 11 at 15:02
  • $\begingroup$ Yes, sorry I didn't put the brackets $\endgroup$ – V11 Mar 11 at 15:02
  • $\begingroup$ Is it possible to get the final part of the equation (part (i)) to y = some value and then replace the x? $\endgroup$ – V11 Mar 11 at 16:23
  • $\begingroup$ Yes, we have $$h^{-1}=\frac{1}{x}$$ so $$h(h^{-1}(x))=2\cdot \frac{1}{2}x=x$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 11 at 16:28

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