Derivatives and Increments -help, again

[Asphyxiated]

Homework Statement

Find f'(c) and the error estimate for:

f(x)= \sqrt{x^{2}+1}

Homework Equations

The error is given by:

E(\Delta x) = \frac {1}{2}M \Delta x

and

f''(c) \leq M

The Attempt at a Solution

So the first derivative is:

f'(x) = \frac {x}{\sqrt{x^{2}+1}}

and for the second derivative I got:

f''(x) = (x^{2}+1)^{-3/2}

So for f'(c) I got:

f'(c) = \frac {2}{\sqrt{5}}

which is correct in the back of the book but for the error i got:

f''(c) = 5^{-3/2}

so

E(\Delta x) = \frac {1}{2} (5)^{-3/2} \Delta x \; \approx \; .04472 \Delta x

but according to the book it is:

E(\Delta x) = \frac {1}{4\sqrt{2}} \Delta x \; \approx \; .17678 \Delta x

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):

f'(x) = \frac {x}{\sqrt{x^{2}+1}}

f''(x) = \frac {(\sqrt{x^{2}+1}) (1) - (x) (\frac {x}{\sqrt{x^{2}+1}})} {(\sqrt{x^{2}+1})^{2}}

which then boils down to:

f''(x) = \frac {\sqrt{x^{2}+1}- \frac {x^{2}}{\sqrt{x^{2}+1}}} {x^{2}+1}

Next let's combine the terms on the top to get something like:

f''(x) = \frac {\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}}}{x^{2}+1}

so x^2 will cancel on the top fraction to get:

f''(x) = \frac {\frac {1}{\sqrt{x^{2}+1}}}{x^{2} +1}

and that can be rewritten as:

\frac {1}{\sqrt{x^{2}+1}} * \frac {1}{x^{2}+1}

which can further be rewritten as:

(x^{2}+1)^{-1/2} * (x^{2}+1)^{-1} = (x^{2}+1)^{-3/2}

so:

f''(x) = (x^{2}+1)^{-3/2}

where did I go wrong?

 

[Mark44]

Where is M? For your error estimate you need a number M such that f''(x) <= M for all x in some interval.

What is c? You show values for f'(c) and f''(c) without ever saying what c is.

f&#039;(c) = \frac {2}{\sqrt{5}}
f&#039;&#039;(c) = 5^{-3/2}

Your result for f''(x) look fine.

 

[Asphyxiated]

Oh my bad, c = 2 and since:

f&#039;&#039;(x) \leq M

f''(c) = M so in this problem (as shown):

M = (x^{2}+1)^{-3/2} \; and \; with \; c=2, \; M=(5)^{-3/2}

 

[Mark44]

There are two parts to this problem:
1. Find f'(2)
2. Find the error estimate for f(x) = sqrt(x^2 + 1).

The first part you have already done.
For the second part, we can estimate f(x) by f(2) + f'(2)\Delta x. The error in this estimate is E(\Delta x), the absolute value of which is equal to (1/2)f''(\xi)\Delta x, for some number \xi between 2 and x.

Since we don't know \xi (the Greek letter xi), the best we can do is find the largest value of f''(\xi), which we'll call M.

As it turns out, the global maximum value for f''(x) = 1/(x^2 + 1)^(3/2) is 1, which is attained when x = 0. Since the answer in your book doesn't seem to be using that value, but seems to be using f''(1) instead, I suspect that the problem is assuming that \Delta x is less than 1.

You neglected to include the fact that c was given as 2. Did you forget to mention another piece of given information?