Intermediate Value Theorem Solution 15

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SOLUTION 15: Consider the graphs of

$y=f(x)$ and

$y=x$ on the interval

$[0, 1]$ with

$f(0)>0$ and

$f(1)<1$ .

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Let function

$g(x)= f(x)-x \ \ \ \ and \ choose \ \ \ \ m=0$
Function

$g$ is continuous on the interval

$[0, 1]$ since it is the DIFFERENCE of continuous functions. Note that

$g(0) = f(0) - 0 = f(0)>0$ and

$g(1)=f(1) - 1 < 0$
i.e.,

$m=0$ is between

$f(0)$ and

$f(1)$ .

The assumptions of the Intermediate Value Theorem have now been met on the interval

$[0, 1]$ , so we can conclude that there is some number

$c$ in the interval

$[0, 1]$ which satisfies

$g(c)=m$ i.e.,

$f(c)-c =0 \ \ \ \ \longrightarrow \ \ \ \ f(c)=c$ This completes the proof.

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