Prove that $\exists c\in(0,1)$ such that $f(c)=c^2$

Question

Suppose that a function $f:[0,1] \to \mathbb{ R}$ is continuous and $3 \int_0^1 f(x)dx = 1$. Prove there exists $c \in (0,1)$ such that $f(c) = c^2 $

please format correctly – dato datuashvili Dec 11 '13 at 13:12 — dato datuashvili, Dec 11 '13 at 13:12

score 4 · Answer 1 · answered Dec 11 '13 at 13:15

4

We have that $f$ is integrable. Let $$h(x)=\int_{0}^{x} f(t)dt-\frac {x^3}{3}$$.

Then $h$ is continuous in $[0,1]$ and differentiable at $(0,1)$ .Also $h(0)=h(1)$. So by Roll'es Theorem we have that there is a $c\in (0,1):h'(c)=0$.

answered Dec 11 '13 at 13:15

Haha

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score 3 · Answer 2 · answered Dec 11 '13 at 13:12

3

Hint $g(x)=\int_{0}^{x}f(t)dt-{x^3\over 3}$, Apply Rolles Thm.

answered Dec 11 '13 at 13:12

Myshkin

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Prove that $\exists c\in(0,1)$ such that $f(c)=c^2$

2 Answers2