Intermediate Value Theorem Solution 13

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SOLUTION 13: We are given the equation

$\tan{x} = 1-x \ \ \ \ \longrightarrow \ \ \ \ \tan{x}+x-1=0$ Let function

$f(x)=\tan{x}+x-1 \ \ \ \ and \ choose \ \ \ \ m=0$ The function

$x-1$ is continuous for all values of

$x$ since it is a polynomial. Consider the fact that

$\tan{x}$ is a well-known periodic function, which is piecewise continuous. Let's look at the graph of both functions to establish an appropriate interval.

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From this graph we can see that the solution lies in the interval

$[0, \pi/4]$ and that

$f$ is continuous on

$[0, \pi/4]$ since it is the SUM of continuous functions.
Note that

$f(0)= \tan{0}+(0)-1= \{0\}-1 =-1 \ \ \ \ and \ \ \ \ f(\pi/4)= \tan{\pi/4}+(\pi/4)-1=1+(\pi/4)-1= \pi/4>0$
so that

$f(0) = -1 < m < \pi/4=f(\pi/4)$
i.e.,

$m=0$ is between

$f(0)$ and

$f(\pi/4)$ . Now choose the interval to be

$\ [0, \pi/4]$ .

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number

$c$ in the interval

$[0, \pi/4]$ which satisfies

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

$\tan{c}+c-1=0$ and the equation is solvable.

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