1
$\begingroup$

We just started doing inverse functions so I'm not very familiar with this concept...

$\endgroup$
1
  • 1
    $\begingroup$ If you can prove that $f(x)$ is one-one and onto, then you are done. Injectivity will follow from the derivative being of the same sign, while surjectivity will follow from continuity, and the fact that $f(x)$ is an odd polynomial. $\endgroup$ Commented Dec 5, 2016 at 3:20

2 Answers 2

3
$\begingroup$

$f'(x)=7x^6+1>0$ implies that $f$ is strictly increasing thus injective. $lim_{x\rightarrow-\infty}f(x)=-\infty$, $lim_{x\rightarrow+\infty}f(x)=+\infty$ implies that $f$ is surjective since for every $n>0$, there exists $a<-n<n<b$ such that $f(x)=a$, $f(y)=b$ and $[a,b]$ is in the image of $f$ since $f$ is continue and the image of $R$ is connected.

$\endgroup$
2
$\begingroup$

$f(x)$ is a continuos function and the derivative $f'(x)=7x^6+1>0$ so the inverse function exists because the function is strictly increasing.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .