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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.

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please use latex –  noname1014 Sep 27 '12 at 4:39
    
When you write c(1,2), I assume you mean $c$ in the open interval $(1,2)$. –  Gerry Myerson Sep 27 '12 at 4:51
    
yes sorry I should make the more clear –  Chance Sep 27 '12 at 5:00

1 Answer 1

up vote 3 down vote accepted

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.

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thank you so much for this answer. But the one thing I can't follow is where the 1/(a-1)>B+2 comes from? –  Chance Sep 27 '12 at 5:13
    
Near $1$ but to the right of $1$, $\frac{1}{a-1}$ will be very large positive, so $\frac{1}{1-x}$ is very large negative, so the term $-\frac{1}{1-x}$ in $g(x)$ is large positive. (Should rewrite, too many minus signs.) We want to make sure that the $f(x)-\frac{1}{2-x}$ term does not spoil things by being too large negative. The $f(x)$ part can't change things by more than $B$, and the $-\frac{1}{2-x}$ term can be controlled by making sure that $x\lt 1.5$, for then it contributes no more than $2$. –  André Nicolas Sep 27 '12 at 5:20
    
wow that is awesome. Everyone I know has been stuck on this problem forever, thanks again! –  Chance Sep 27 '12 at 5:25
    
@Chance: I changed the writeup a bit: In the last part I replaced $-\frac{1}{1-x}$ by $\frac{1}{x-1}$. Then we don't end up chasing negatives of negatives. –  André Nicolas Sep 27 '12 at 5:32

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