# Find condition for one-one ness

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?

HINT: You need $$f'(x)=1+ag'(x)\ne 0$$. Can you proceed from here?
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.
• 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. Aug 22, 2019 at 3:48