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Let $g:\mathbb{R} \to \mathbb{R}$ be a differentiable function such that $|g'(x) |\le M$ for all $x\in \mathbb{R}$. For what values of $a$ will the function $f(x) = x + a g(x)$ be necessarily one-to-one?

As $g'$ is bounded $g$ is uniformly continuous. Hence $f$ is continuous. Now I tried to use the result that If a continuous function is monotone then it is one one. But I failed to get any conclusion. What was my fault? What is the right way?

desiigner
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Soumyadweep Mondal
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2 Answers2

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You are on the right track.

HINT: You need $f'(x)=1+ag'(x)\ne 0$. Can you proceed from here?

Nitin Uniyal
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The fact that $g'$ exists is enough to tell you that $g$ is continuous and therefore that $f$ is continuous. Now if $|a|<1/M$ then:

$$f'(x)=\frac{\text{d}}{\text {dx}}(x+ag(x))=1+ag'(x)> 1-a\frac{1}{a}=0$$ so f is monotone increasing, therefore one-to-one.

subrosar
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  • Sorry I wrote this before I saw the hint, and I thought it was important to note that the slope should be positive, since a bound on $a$ cannot guarantee the slope is negative. – subrosar Aug 22 '19 at 03:48