Barrett Osborn

2022-11-23

Use the intermediate value theorem to prove that if $f:\left[1,2\right]\to R$ is a continuous function, that there is at least one number $c$ in the interval $\left(1,2\right)$ such that $f\left(c\right)=1/\left(1-c\right)+1/\left(2-c\right)$
This is a question for my intro calc class that I am having a hard time understanding.

Kailee Abbott

Expert

Note that because $f\left(x\right)$ is continuous on $\left[1,2\right]$, the function $f\left(x\right)$ is bounded on $\left[1,2\right]$. Suppose that $|f\left(x\right)| for all $x$ in our interval. Let
$g\left(x\right)=f\left(x\right)-\frac{1}{1-x}-\frac{1}{2-x}.$
There is an $a$ in (1,2) such that $g\left(a\right)$ is positive, and a $b$ such that $g\left(b\right)$ is negative, and hence by the Intermediate Value Theorem there is a $c$ between $a$ and $b$ such that $g\left(c\right)=0$.

Detail: We show that there is indeed an $a$ such that $g\left(a\right)$ is positive.
In order to have fewer minus signs, note that
$g\left(x\right)=f\left(x\right)+\frac{1}{x-1}-\frac{1}{2-x}.$
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 $\frac{1}{a-1}>B+2$, we can make sure that $g\left(a\right)$ is positive. For then the $f\left(a\right)-\frac{1}{2-a}$ part cannot be negative enough to make $g\left(a\right)$ 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 $g\left(x\right)$ is large negative.

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