3
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ thanks that solved the problem. $\endgroup$ – rbtLong Feb 1 '13 at 7:26
  • $\begingroup$ 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. $\endgroup$ – rbtLong Feb 1 '13 at 8:04
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – rbtLong Feb 1 '13 at 7:27
  • $\begingroup$ No; my hint is essentially the same as André's. I was suggesting how you could write $x^2-5$ as a function of $x-5$. $\endgroup$ – Zev Chonoles Feb 1 '13 at 7:30
  • $\begingroup$ ah ok thanks for the answer! $\endgroup$ – rbtLong Feb 1 '13 at 7:30
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – André Nicolas Feb 1 '13 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.