Prove that a function is injective knowing the second derivative is positive

Question

I have this problem and I'm trying to figure out how to start:

Suppose that $f$ has positive second derivative in $(a,b)$, then prove that $f$ is injective in $(a,b).$

I manage to show that $f'(x)$ is injective. Let $x_1,x_2\in(a,b),$ with $x_1≠x_2.$ By the mean value theorem, taking the interval $(x_1,x_2)$ exist $\xi\in(x_1,x_2)$ such that: $$\frac{f'(x_2)-f'(x_1)}{x_2-x_1}=f''(\xi);$$ $$f'(x_2)-f'(x_1)=(x_2-x_1) f''(\xi).$$ We have that $f''(\xi)\gt0$ by hypothesis and also $x_2-x_1\gt0,$ assuming that $x_2\gt x_1$ without losing generallity. So $f'(x_2)-f'(x_1)\gt0$ and that implies that $f'(x_1)≠f'(x_2).$ Anyone know how to proceed? Thanks.

Answer 1

This is false. You can prove it is injective (even increasing) its first derivative is positive, or that it is convex if its second derivative is positive, but an easy counterexample to the posted claim can be found above.

Answer 2

If second derivative of a function is positive on an interval, then the function is concave up but it does not have to be injective.

For example the function $$f(x)= x^2+1$$ on the interval $(-2,2)$ is not injective because $$f(1)=f(-1) = 2 $$ while its second derivative is $ f''(x) =2 >0$

Answer 3

As they said is false. Also you can do a similar proof of this similar proposition: If $f'(x)≠0$ for all $x$, then $f$ is injective