Barrett Osborn

Answered

2022-11-23

Use the intermediate value theorem to prove that if $f:[1,2]\to R$ 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.

This is a question for my intro calc class that I am having a hard time understanding.

Answer & Explanation

Kailee Abbott

Expert

2022-11-24Added 14 answers

Note that because $f(x)$ is continuous on $[1,2]$, the function $f(x)$ is bounded on $[1,2]$. Suppose that $|f(x)|<B$ 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 $\frac{1}{a-1}>B+2$, 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.

$g(x)=f(x)-\frac{1}{1-x}-\frac{1}{2-x}.$

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 $\frac{1}{a-1}>B+2$, 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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