3
$\begingroup$

How do I go about solving a problem like this? Do I find the inverse formula first? If so I am stuck at $\frac{x}{y} = y^3(y^2+1)$. Help is welcomed.

$\endgroup$
1
  • $\begingroup$ It's great if you can find a general inverse formula, but it will be very hard to do so in this case. So your efforts are best focused on looking for values of $x$ where $f(x) = 3$, and then trying to decide if there are any other values of $x$. $\endgroup$
    – Erick Wong
    Jun 23 '15 at 18:38
3
$\begingroup$

Note that for the second one, $f$ and $f^{-1}$ essentially cancel. So $f(f^{-1}(2))=2$.

For the former, letting $b=f^{-1}(3)$, we have $f(b)=f(f^{-1}(3))=3$. So we need to find $b$ such that $f(b)=3$. Then

$$b^5+b^3+b=3$$

which I think has at least one pretty clear solution.

$\endgroup$
2
$\begingroup$

Given the phrasing of the question, it's unlikely that you are expected to find the inverse; you have been asked a question to which the answer is very guessable (in particular, $f^{-1}(3)=1$), and for the second, we know that the composition of a function with its inverse is the identity. The question seems to be checking your understanding of the notion of an inverse function, so don't overthink it!

However, you do need to check or show that the inverse for the former is unique!

$\endgroup$
2
$\begingroup$

OK, I admit I didn't see this at first, but here's a hint:

If you're looking for $f^{-1}(3)$ then you want to solve this equation for $x$:

$$3 = x^5 + x^3 + x.$$

Can you see a solution for $x$ by inspection?

$\endgroup$
1
$\begingroup$

The continuos function $f:\mathbb{R} \to \mathbb{R}$ defined by rule $f(x)=x^5+x^3+x$ is stricly incrasing, then the equation $f(x)=3$ has only a single solution. But $1^5+1^3+1=3$, then $x=1$ is single solution the equation $f(x)=3$ therefore $f^{-1}(3)=1$. Clearly $f^{-1}(f(2))=2$ because $f$ is bijection.

$\endgroup$

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.