# If derivative of function is less than or equal to 4, prove the existence of point $c$ so $f(c)=c^2$.

I am struggling to construct a simple proof of the following.

Suppose $f(x)$ is differentiable for any real value $x$ and that $f'(x) \le 4$ for all real values of x. Prove there is at most one real value $c>2$ such that $f(c) =c^2$.

I am completely lost on where to start on this proof and any help would be greatly appreciated!

• Your heading doesn't match the text. Are you trying to prove existence or uniqueness or both? – lulu Dec 11 '15 at 0:25
• Note: the function $f(x)=x$ is a counterexample to the text question (as $f(0)=0^2$ and $f(1)=1^2$). Maybe the text was meant to read "at LEAST one real value..." – lulu Dec 11 '15 at 0:28
• @lulu $0$ and $1$ are not values satisfying $c>2$. – Clement C. Dec 11 '15 at 0:29
• @ClementC. Ah, missed that condition. Thanks! – lulu Dec 11 '15 at 0:30
• Assume that there are more than $1$ value of such $c$. Then they both lie on the curve $y=x^2$ for $x>2$, and in particular since $x^2$ is convex the gradient of line connecting those two points is greater than $4$. Then use MVT on $f$ for these two points to show that there exists a point $x_0$ between these two points such that $f'(x_0)>4$. – user160738 Dec 11 '15 at 0:32

Say there were two such values $f(a)=a^2$, $f(b)=b^2$ with $a,b>2$ (we will obtain a contradiction) Then the slope of the chord connecting those two points on the graph is $$m=\frac {b^2-a^2}{b-a}=b+a$$
As $a,b>2$ we see that $m>4$. Now, the MVT tells us that there is some value $t\in [a,b]$ with $f'(t)=m>4$. But this contradicts our assumption on the derivative, so we are done.
• Sorry, not following. I was working by contradiction. I assumed (with intent of disproving) that there were two such points. I then derived a contradiction from that assumption (specifically, I used the two solutions plus the MVT to show that the derivative had to be greater than $4$ somewhere). – lulu Dec 11 '15 at 1:02