For a continuous function $f$ show $\exists c\in (0,1)$ s.t $f(c)=3c^2$

Let $f: [0,1] \to \mathbb{R}$ be a continuous function such that $$\int_0^1 f (t) dt =1 .$$ Show that there exists a point $c \in (0,1)$ such that $f(c)=3c^2$.

I tried using mean value theorem of integrals on $f (x)$, so $$\int_0^1 f (t) dt = f (c) =1,$$ but it should be $3 c^2$. What is going wrong?

HINT: Let $g(x) = \int_{0}^{x}f(t)\,dt - x^{3}$. Then $g(0) = g(1) = 0$. Now apply Rolle's Theorem on $g(x)$.

• What made u think of rolls theorm here – ketan Jan 13 '15 at 4:42
• @ketan: From the question you should see that $f(c) - 3c^{2} = 0$ and hence we need a function whose derivative is $f(x) - 3x^{2}$ and this derivative needs to vanish at $c$. Hence I construct $g(x)$ whose derivative is as desired. Now for Rolle's we must have $g(0) = g(1)$ which is the case here and then $g'(c) = 0$ for some $c \in (0, 1)$. In most such problems you use Rolle's theorem. – Paramanand Singh Jan 13 '15 at 4:45
• Very very nicely explained thank you – ketan Jan 13 '15 at 4:47
• @ketan: BTW thanks to you too for accepting this answer. In such problems the real challenge is to find/construct a suitable function $g(x)$. You will find many such questions here on MSE. – Paramanand Singh Jan 13 '15 at 4:49

Here is another way. Since $\int_0^1 f(t)\, dt = 1 = \int_0^1 3t^2\, dt$, we have $\int_0^1 (f(t) - 3t^2)\, dt = 0$. By the mean value theorem for integrals, there is a $c\in (0,1)$ such that $f(c) - 3c^2 = \int_0^1 (f(t) - 3t^2)\, dt$, i.e., $f(c) - 3c^2 = 0$, or, $f(c) = 3c^2$.

If $f(t)> 3t^2$ for all $t\in [0,1]$ then $\int_0^1 f(t) dt > \int_0^1 3 t^2 dt= 1$, contradiction. Similarly, it is not possible that $f(t) < 3 t^2$ for all $t \in [0,1]$.

Therefore, we can find two points $t_1$, $t_2$ so that $f(t_1) \le 3 t_1^2$ and $f(t_2)\ge 3 t_2^2$. Between them lies an $s$ so that $f(s) = 3 s^2$.

Note that we used the continuity of $f$. The whole point with $3t^2$ is that $\int_0^1 3t^2 dt= 1$, the value imposed on $\int_0^1 f(t) dt$.