Derivatives and Increments -help

[Asphyxiated]

Homework Statement

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

$$f'(c) = \lim_{x \to 0^{+}} \frac {f(c+\Delta x) - f(c)}{\Delta x}$$

I just included that to show that we are working with one (right) sided limit the actual problem is:

$$f(x) = \frac {1}{x} \;\; with\;\; c = 3$$

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 this really quite simple, if:

$$f(x)= \frac {1}{x}$$

then

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

and

$$f''(x) = \frac {2}{x^{3}}$$

so

$$f'(c) = -\frac{1}{9}$$

and

$$f''(c) = \frac {2}{27} \Delta x$$

but according to the book the error is suppose to be 1/27 which doesn't really make sense to me. I got f''(x) like so:

$$\frac {(x^{2})(0)-(-1)(2x)}{(x^{2})^{2}}$$

$$\frac {2x}{x^{4}}$$

$$\frac {2}{x^{3}}$$

where did I go wrong?

 

[Mark44]

Homework Statement

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

$$f'(c) = \lim_{x \to 0^{+}} \frac {f(c+\Delta x) - f(c)}{\Delta x}$$

I just included that to show that we are working with one (right) sided limit the actual problem is:

$$f(x) = \frac {1}{x} \;\; with\;\; c = 3$$

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 this really quite simple, if:

$$f(x)= \frac {1}{x}$$

then

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

and

$$f''(x) = \frac {2}{x^{3}}$$

so

$$f'(c) = -\frac{1}{9}$$

and

$$f''(c) = \frac {2}{27} \Delta x$$

but according to the book the error is suppose to be 1/27 which doesn't really make sense to me. I got f''(x) like so:

$$\frac {(x^{2})(0)-(-1)(2x)}{(x^{2})^{2}}$$

$$\frac {2x}{x^{4}}$$

$$\frac {2}{x^{3}}$$

where did I go wrong?

What are you using for $\Delta x$? I looked pretty closely and didn't see it anywhere.

 

[Asphyxiated]

Ah yeah, that is probably a little confusing, right now $$\; \Delta x \;$$ does not have a value as the first problems of the chapter only ask me to find f'(c) and the error. $$\Delta x \;$$ will only take on a value in the later problems which as for an estimated value for something like:

$$\sqrt{65}$$

 

[Mark44]

I guess you could take $\Delta x$ to be 1, so that x is in the interval [3, 4]. (You said in your OP that you were interested in the right-hand limit.) In this interval, the largest value of f''(x) = 2/x³ is 2/27, which occurs when x = 3. This is because f''(x) is a decreasing function on this interal.

Then E($\Delta x$) = (1/2)M $\Delta x$ = ?

 

[Asphyxiated]

Oh ok I see now mark44, I simply forgot to times M by 1/2 so yeah:

$$E(\Delta x) = \frac {1}{2} \frac{2}{27} \Delta x = \frac {1}{27} \Delta x$$

Note that $$\; \Delta x \;$$ is not 1, it does not have a value at this point in the exercises. The correct answer in the back is written as:

$$f'(c) = -\frac {1}{9},\;\;\; error= \frac {1}{27} \Delta x$$

Thanks, I just needed to pay more attention.

 

[Mark44]

Also, you said in the first post that the error was given in the back of the book as 1/27, rather than 1/27 * delta x. That threw me off.

 

[Asphyxiated]

Yeah I realized that... I apologize for any confusion do to that.