Let f be differentiable, with a positive derivative. Show that f is onto.

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable with $f'(x) \geq 1, \forall x \in \mathbb{R} $. Show $f$ is onto.

I was wondering if I could have a hint, and then work through this problem with people.

I tried starting with the definition of a derivative at a point, but I'm not sure I follow how to show it's onto.

Thank you,

I do understand the reasoning that because the derivative is positive, then the function must necessarily always be increasing. A monotonically increasing function from the reals, must be onto.

But I'm not sure how to show that.

Answer 1

Hint: Try to show that $f(x) \geq x + f(0)$ and then apply the intermediate value theorem.

Answer 2

Hint:

MVT:

Suppose $f$ continuous on $[a,b$] , $f$ differentiable differentiable in $(a,b)$, then there is a point $t \in (a,b)$ with

$\dfrac{f(b)-f(a)}{b-a}=f'(t) \ge 1.$

1) Choose $x=b, a=0$, for example, and cosider $x \rightarrow \infty.$

2) Similarly : $x \rightarrow -\infty.$

What can you say about f(x) ?

Answer 3

By continuity, the image of $f$ will be an interval. Suppose that this interval is $\ne\Bbb R$. Suppose wlog that the interval is bounded above and let be $M = \sup\{f(x) : x\in\Bbb R\} < +\infty$. Take $x_0\in\Bbb R$ s.t. $f(x_0) > M - 1$. As $f$ is increasing. $$M - 1 < f(x_0) < f(x_0 + 1)\le M.$$ Now, apply the mean value theorem in $[x_0,x_0 + 1]$.