Using the intermediate value theorem to prove that there is at least one number $c\in(1,2)$ such the $f(c)=1/(1-c)+1/(2-c)$ Use the intermediate value theorem to prove that if $f:[1,2]\rightarrow R$ is a continuous function, that there is at least one number $c$ in the interval $(1,2)$ such that $f(c)=1/(1-c)+1/(2-c)$
This is a question for my intro calc class that I am having a hard time understanding.
 A: Note that because $f(x)$ is continuous on $[1,2]$, the function $f(x)$ is bounded on $[1,2]$. Suppose that $|f(x)|\lt B$ for all $x$ in our interval. Let
$$g(x)=f(x)-\frac{1}{1-x}-\frac{1}{2-x}.$$
There is an $a$ in $(1,2)$ such that $g(a)$ is positive, and a $b$ such that $g(b)$ is negative, and hence by the Intermediate Value Theorem there is a $c$ between $a$ and $b$ such that $g(c)=0$.
Detail: We show that there is indeed an $a$ such that $g(a)$ is positive.
In order to have fewer minus signs, note that
$$g(x)=f(x)+\frac{1}{x-1}-\frac{1}{2-x}.$$
Note that $\frac{1}{x-1}$ becomes very large positive for $x$ close enough to $1$ but to the right of $1$. 
The term  $\frac{1}{2-x}$ is close to $1$ when $x$ is close to $1$. So by choosing $a$ near $1$ such that $\frac{1}{a-1}\gt B+2$, we can make sure that $g(a)$ is positive. For then the $f(a)-\frac{1}{2-a}$ part cannot be negative enough to make $g(a)$ negative. 
For $b$, we play the same game near $2$. For $x$ near $2$ but to the left of $2$, the term $\frac{1}{2-x}$ is large positive, so $g(x)$ is large negative.
