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Homework Statement
Find f'(c) and the error estimate for:
[tex] f(x)= \sqrt{x^{2}+1} [/tex]
Homework Equations
The error is given by:
[tex] E(\Delta x) = \frac {1}{2}M \Delta x [/tex]
and
[tex] f''(c) \leq M [/tex]
The Attempt at a Solution
So the first derivative is:
[tex] f'(x) = \frac {x}{\sqrt{x^{2}+1}} [/tex]
and for the second derivative I got:
[tex] f''(x) = (x^{2}+1)^{-3/2} [/tex]
So for f'(c) I got:
[tex] f'(c) = \frac {2}{\sqrt{5}} [/tex]
which is correct in the back of the book but for the error i got:
[tex] f''(c) = 5^{-3/2} [/tex]
so
[tex] E(\Delta x) = \frac {1}{2} (5)^{-3/2} \Delta x \; \approx \; .04472 \Delta x[/tex]
but according to the book it is:
[tex] E(\Delta x) = \frac {1}{4\sqrt{2}} \Delta x \; \approx \; .17678 \Delta x [/tex]
I have done the second derivative over now a couple of times and still get the same answer but here is how i got f''(x):
[tex] f'(x) = \frac {x}{\sqrt{x^{2}+1}} [/tex]
[tex] f''(x) = \frac {(\sqrt{x^{2}+1}) (1) - (x) (\frac {x}{\sqrt{x^{2}+1}})} {(\sqrt{x^{2}+1})^{2}} [/tex]
which then boils down to:
[tex] f''(x) = \frac {\sqrt{x^{2}+1}- \frac {x^{2}}{\sqrt{x^{2}+1}}} {x^{2}+1} [/tex]
Next let's combine the terms on the top to get something like:
[tex] f''(x) = \frac {\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}}}{x^{2}+1} [/tex]
so x^2 will cancel on the top fraction to get:
[tex] f''(x) = \frac {\frac {1}{\sqrt{x^{2}+1}}}{x^{2} +1} [/tex]
and that can be rewritten as:
[tex] \frac {1}{\sqrt{x^{2}+1}} * \frac {1}{x^{2}+1} [/tex]
which can further be rewritten as:
[tex] (x^{2}+1)^{-1/2} * (x^{2}+1)^{-1} = (x^{2}+1)^{-3/2} [/tex]
so:
[tex] f''(x) = (x^{2}+1)^{-3/2} [/tex]
where did I go wrong?