Continuity and integration without given functions

Question

I have a continuous function $f:[0,1]\rightarrow\mathbb{R}$ and a function $F$ for which is true that:

• $F'(x)=f(x)$
• $F(1)-F(0)=1$

Now I want to prove that:

• exists $x_1\in(0,1)$ in order to $f(x_1)=2x_1$
• exists $x_2\in(0,1)$ in order to: $$f(x_2)=\frac{F(1)-1-F(x_2)}{x_2-1}$$

For the first part I am thinking of using Bolzano's theorem, but I don't know how I can pull this off and for the second part I don't have a lot of ideas.

Any help?

Answer 1

Hint. Use the Mean Value Theorem.

For i) consider $g(x)=F(x)-x^2$. Then $g(0)=F(0)$ and $g(1)=F(1)-1=F(0)$.

For ii) consider $h(x)=F(x)(x-1)$. Then $h(0)=-F(0)=1-F(1)$ and $h(1)=0$.

Answer 2

Since, $F'$ exists, I take it we may assume that $F$ is differentiable?

Then by the Fundamental Theorem of Calculus (FTC), we have that \begin{align} \int _0^1 F'(x)dx & =F(1)-F(0)\\ \implies \int _0^1 f(x)dx &=1 \end{align}

Now consider the function $g(x)=f(x) -2x$ defined on the interval $[0,1]$, and observe that $$\int _0^1 g(x)dx=\int _0^1 f(x)dx-\int _0^1 2xdx=1-1=0$$

Suppose that $g(x)<0 \; \forall \; x \in [0,1]$. Then $\int _0^1 g(x)dx<0$, contradiction.

Similarly, $g(x)<0$ on all of $[0,1]$ also gives rise to contradiction.

Thus, there must exist $a,b \in [0,1]$ such that $g(a)<0$ and $g(b)>0$, and by the Intermediate Value Theorem (IVT), there exists $x_1$ between $a$ and $b$ such that $g(x_1)=0$

Finally, $g(x_1)=0 \implies f(x_1)-2(x_1)=0 \implies f(x_1)=2x$