Let $f(x)$ be defined on $[a,b]$ such that $$f(x)=x+\frac{g(x)}{20}$$ Where $g(x)$ is differentiable function on [a,b] and $|g'(x)|\leq 10$, Then:

a) $f(x)$ is bounded variation.( I have verified this, because the function has bounded derivative)

b) $f(x)$ is one one.

Please tell me how can prove the function is one one.

  • $\begingroup$ Down voter should leave a comment so that I can improve my questions in future. $\endgroup$ – Rayees Ahmad Mar 30 '16 at 7:18
  • 1
    $\begingroup$ although I didn't down vote, some person probably did because of your previous title $\endgroup$ – Nikunj Mar 30 '16 at 7:18
  • $\begingroup$ I'm upvoting to cancel the anonymous downvote. Don't see anything wrong with the question/ $\endgroup$ – Shailesh Mar 30 '16 at 7:22

Fill in details:

$$f'(x)=1+\frac{g'(x)}{20}\stackrel{\text{Why? Justify}}>0\implies\;f\;\;\text{is injective}$$

since it is monotonic increasing in the given interval

  • 1
    $\begingroup$ *monotonic increasing? $\endgroup$ – G. Bach Mar 30 '16 at 12:38
  • $\begingroup$ @G.Bach Thank you. Of course, typo corrected. $\endgroup$ – DonAntonio Mar 30 '16 at 14:02

A sufficient condition for $f$ to be injective ("one-to-one") is for $f'$ to be either strictly positive, or strictly negative. In your case, since $-10 \le g' \le 10$, and $f' = 1 + \frac {g'} {20}$, it follows that

$$\frac 1 2 = 1 + \frac {-10} {20} \le \underbrace {1 + \frac {g'} {20}} _{f'} \le 1 + \frac {10} {20} = \frac 3 2 ,$$

so $f'(x) \in [\frac 1 2, \frac 3 2] \ \forall x \in [a,b]$, which shows that $f'$ is strictly positive, therefore strictly increasing and thus injective.


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.