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=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#.