On Problem Set 5, I don't part of the answer to 2G-2b. How do they arrive at (1+x)^(1/2) = 1+(1/(2(1+c)^(1/2))*x ? (it's the second to last step. I know the form I've presented there is hard to read.)

I solved the problem a different way, but I don't understand their solution.

Question

On Problem Set 5, I don't part of the answer to 2G-2b. How do they arrive at (1+x)^(1/2) = 1+(1/(2(1+c)^(1/2))*x ? (it's the second to last step. I know the form I've presented there is hard to read.)

I solved the problem a different way, but I don't understand their solution.

anonymous · Answer

Here's the form of the MVT they're using:$$f(b)=f(a)+f'(c)(b-a)$$In the solution, a=0 because the problem asks about the situation when x>0. We're also going to insert x in place of b. We get the following equivalencies:$$f(b)=\sqrt{1+x}$$$$f(a)=\sqrt{1+0}=1$$$$f'(c)=\frac{ 1 }{ 2\sqrt{1+c} }$$$$b-a=x-0=x$$Plugging all these into the MVT formula produces what you see in the printed solution.