$f:[a,b]\to\mathbb{R}$ continuous function and differentiable in (a,b) $\rightarrow$ $\exists$ $c\in(a,b)$ with $f(c)=\frac{1}{a-c}+\frac{1}{b-c}$

Question

I am preparing for my exam and need help for the following task:

Let $f:[a,b]\to\mathbb{R}$ be a continuous function and differentiable in (a,b).

a) Show that $c \in(a,b)$ exist such that $f(c)=\frac{1}{a-c}+\frac{1}{b-c}$

Well I thought a lot about this task. I thought maybe we could use the intermediate value theorem and use the function $f(x)=\frac{1}{a-x}+\frac{1}{b-x}$. Since f is continuous in [a,b] we must have that $$\lim\limits_{x \uparrow a}{f(x)}=f(a)$$ But this limit does not exist, so we can't work with $f(x)=\frac{1}{a-x}+\frac{1}{b-x}$.

So we don't have an explicit function to work with. What could we use. We can't use intermediate value theorem. Also we don't know anything about f(a) and f(b). If we would know that f(a)<f(c)<f(b), then we could prove this easily. Maybe our knowledge about the fact, that a continuous function $f:[a,b]\to\mathbb{R}$ takes on its absolute maxima and minima could help us? I really don't have any idea how to solve this. Same goes for the next task:

b) Show that there exist $d\in(a,b)$ such that $$f'(d)=\frac{1}{a-d}+\frac{1}{b-d}$$ Hint:Look at the function F(x)=$e^{f(x)}$ (x-a)(x-b)

I thought we could do this like that: Since F(a)=F(b)=0 we could use rolle's theorem. So there exists a $z\in(a,b)$ with F'(z)=0. Which would mean that f(z)=0. But I guess its the wrong way since I don't have any idea what to start with this method.

Is there anyone who could give me an advice. I would be very thankful.

Answer 1

Hints:

1. Consider $f(x)(a-x)(b-x)-(a+b-2x)$ This function has the value $a-b$ at $x=a$ and $b-a$ at $x=b$. Since these have opposite signs there must be some point $c$ at which this function vanishes.

2. $e^{f(x)}(x-a)(x-b)$ vanishes at $x=a$ and $x=b$. Hence, its derivative vanishes at some point $d$. Write down the derivative to complete 2).

Answer 2

I'm going to denote the function $\frac1{a-x}+\frac1{b-x}$ by $g(x)$ to make it clearly different from the function $f()$ we're given in the conditions of the theorem. Here's a hint: since $f()$ is continuous and differentiable, then it takes minimum and maximum values $m$ and $M$ on $[a,b]$. Can you see how to use this to guarantee a value of $x$ close to $a$ such that $f(x)-g(x)\gt 0$, and a value of $x$ close to $b$ such that $f(x)-g(x)\lt 0$?