Is function has exactly one real root? If a function is injective, continuous and differentiable everywhere.

Question

If a function is injective, continuous, and differentiable everywhere. Can we directly say the function has exactly one root? It seems function has exactly one root If a function is injective, continuous, and differentiable everywhere. I think It's the intermediate value theorem. But I think this injective graph hasn't a root?

when the graph goes minus infinity actually can we say the gradient is zero???
I haven't the tough idea If a function is injective, continuous, and differentiable everywhere, It has exactly real root.
Can you offer some assistance to clarify this mess.
Sorry for the poor English language manipulation
Thank you!

Answer 1

The function $f(x)=1+e^x$ is injective, continuous, and differentiable everywhere, yet it has no roots.

Answer 2

You're correct that these conditions don't guarantee a real root. An easy example is $y=e^x$, which is injective, continuous, and differentiable everywhere, but has no real roots. Of course, any injective function can only have at most $1$ real root.