In the first Mean Value Theorem #f(b)=f(a)+(b-a)f'(c), a<c<b, f(x) =log_2 x, a=1 and f'(c)=1. How do you find b and c?

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Answer 1 · 2016-11-06T21:17:29

$b = 1, 2; c = \frac{1}{ln 2}$ $b=1,2;c=1/ln2$

$f(b)=f(a)+(b-a)f'(c)$

$f(x)=log_2x$

Note that if $a=1$ , then $f(a)=log_2 1=0$ .

Also note that $f(x)=lnx/ln2$ so $f'(x)=1/(xln2)$ .

If $f'(c)=1$ , then $1/(cln2)=1$ so $c=1/ln2$ .

Now we can also solve for $b$ :

$f(b)=f(a)+(b-a)f'(c)$

We can replace $f(b)$ with $log_2b$ , $f(a)$ with $0$ , $a$ with $1$ , and $f'(c)$ with $1$ .

$log_2b=0+(b-1)(1)$

$log_2b=b-1$

This can be manipulated to show:

$b=2^(b-1)$

Or:

$2b=2^b$

This occurs at $b=1$ and $b=2$ .