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so this is a very simple question but I am having a tough time with it.

So it's finals week and I'm studying up for an Algebra 2 final. The only part I am having trouble with is finding the inverse of a function. It doesn't confuse me that I have to basically undo everything that is happening to x (or any variable) but it confuses me what order I have to undo everything in.

Say I have the expression $f(x)= \sqrt{x-3}$. I would first substitute $f(x)$ with y and then switch the x and y so I'd be finding x. After that I'd square x and then add 3 to x to make the equation $y = x^2 + 3$.

But what if I had the expression $f(x)= \sqrt{x-3} + 5$? Would I subtract 5 and then square the x and 5 or would I square x and then add 3 and subtract 5?

Thanks.

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2 Answers 2

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I think you're on the right track, but maybe a bit muddled in your thought; yes, the way to invert the function $f(x)=\sqrt{x-3}+5$ is to start with $y$, subtract $5$ from it, square that, then add $3$. That is, $$(y-5)^2+3=x.$$

The easiest way to think about this is to consider what such a function is doing; that is, $f(x)$ could be described as:

  • Start with $x$. Subtract $3$ from it.
  • Take the square root of the result of that.
  • Add five to that result.

Symbolically, we could describe this as a composition of functions - that is, if we let $$a(x)=x-3$$ $$b(x)=\sqrt{x}$$ $$c(x)=x+5$$ then we can describe $$f(x)=c(b(a(x))).$$

How do we undo this? Well, we undo each step, in reverse order. That is,

  • To undo the last step, we subtract $5$ from $y$.

  • Then, to undo the step before that, we square that value.

  • Finally, to undo the first step, we add $3$ to the result.

    Symbolically, this is to say that $$f^{-1}(y)=a^{-1}(b^{-1}(c^{-1}(y)))$$ where $$c^{-1}(x)=x-5$$ $$b^{-1}(x)=x^2$$ $$a^{-1}(x)=x+3.$$

That is, we invert each of the "component" function, and then we just plug them in in the reverse order. Then, of course this works because we'd have $$f^{-1}(f(x))=a^{-1}(b^{-1}(c^{-1}(c(b(a(x))))))$$ but we can cancel the inner $c^{-1}(c(\text{stuff}))$ to just the stuff inside, as $c^{-1}$ undoes $c$ and we keep canceling like that to get that $f^{-1}(f(x))=x$, as desired.

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  • $\begingroup$ Wow, thanks. I have never seen it explained that way and it makes much more sense. $\endgroup$
    – Locke
    Jan 22, 2015 at 4:09
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Use the order of operations to track what happens to $x$. In your example, $f(x) = \sqrt{x-3} + 5$, we have (starting with $x$):

  • subtract $3$

  • take the (positive) square root

  • add $5$.

To undo these and solve for $x$, we would undo each, in reverse order. (You noted this much in your question).

  • subtract $5$

  • square

  • add $3$

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