# How can I prove that the following function is one-one?

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.

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

• *monotonic increasing? – G. Bach Mar 30 '16 at 12:38
• @G.Bach Thank you. Of course, typo corrected. – 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.