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Let $f(x) = x-5$, $g(x) = x^2 -5$. Find $u(x)$ if $(u \circ f)(x) = g(x)$.

I know how to do it we have $(f \circ u) (x)$, but only because $f(x)$ was defined. But here $u(x)$ is not defined. Is there any way I can reverse it to get $u(x)$ alone?

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up vote 4 down vote accepted

Hint: We are told that $u(x-5)=x^2-5$. Let $t=x-5$. Now express $x^2-5$ in terms of $t$.

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thanks that solved the problem. – rbtLong Feb 1 '13 at 7:26
can you please check my solution (at the bottom)? the teacher solved it, but i'm not sure if i copied it right. i just dont want to post another question. – rbtLong Feb 1 '13 at 8:04

Hint: $$x^2-5=(x-5)(x+5)$$ $$(x+5)=(x-5)+10$$

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are u using f inverse? my professor used an f inverse on the left and f on the right (this confused me) is that what you're doing? – rbtLong Feb 1 '13 at 7:27
No; my hint is essentially the same as André's. I was suggesting how you could write $x^2-5$ as a function of $x-5$. – Zev Chonoles Feb 1 '13 at 7:30
ah ok thanks for the answer! – rbtLong Feb 1 '13 at 7:30

I think I figured out what my professor did now . . .

$(u \circ f)(x) = g(x)$

$(u \circ f)(f^{-1} (x)) = g( f^{-1}(x)) $

$\big((u \circ f) \circ f^{-1}\big)(x) = (g \circ f^{-1})(x) $

$\big(u \circ (f \circ f^{-1})\big)(x) = (g \circ f^{-1})(x) $

$u(x) = g(f^{-1}(x))$

$u(x) = g(x+5)$

I think this is right. Please correct me if I'm wrong.

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It is essentially correct. There should be a line after the first that says $(u\circ f)(f^{-1}(x))=g(f^{-1}(x))$. Then the rest follows, and is fine except for some missing parentheses. The only downside to this approach is that it is very manipulational, and perhaps gives a little less concrete knowledge of what is going on. – André Nicolas Feb 1 '13 at 8:06

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