0
$\begingroup$

My book gives a few definitions/formulas for obtaining linear approximation, but I'm having trouble understanding how to use them.

Heres the question:

a.) Use the Linear Approximation for f(x) = ln($x^2$ +3) at a = 1 with $\Delta x$ = 0.7 to approximate $\Delta f \approx$ ______________

b.) That Linear Approximation to the value of f(a + $\Delta x$) has error __________ (use calculator to get decimal value)

c.)That Linear Approximation in f(a + $\Delta x$) has percentage error __________

My book says something like this: $\Delta f$ = f(a + $\Delta x$) - f(a), where $\Delta x$ is small. By definition the derivative is the limit.

f '(a) = $\frac{lim}{\Delta x\rightarrow0}\Delta f$ = $\frac{lim}{\Delta x\rightarrow0}(\frac{f(a + \Delta x) - f(a)}{\Delta x})$ = $\frac{lim}{\Delta x\rightarrow0}\frac{\Delta f}{\Delta x}$ So when $\Delta x$ is small, we have $\frac{\Delta f}{\Delta x}\approx$ f '(a) and thus, $\Delta f \approx$ f '(a) $\Delta x$

I thought I was applying the formula correctly, but I guess not. Here's what I've done:

my work

Can someone please help me out with solving this 3 part question? Thanks.

$\endgroup$
1
$\begingroup$

I would have thought (a) required $\Delta f = f(a + \Delta x) - f(a) \approx f'(a)\; \Delta x$ and asking for the last of these expressions

so (b) is $f(a + \Delta x) - (f(a) + f'(a)\; \Delta x)$

and (c) might be $\dfrac{ f(a + \Delta x) - (f(a) + f'(a)\; \Delta x)}{f(a + \Delta x) }$ perhaps multiplying by $100$, though the desired denominator might be $\Delta f = f(a + \Delta x) - f(a)$ rather than $f(a + \Delta x)$.

$\endgroup$
  • $\begingroup$ Ok so for b it would be ln(5.89) $\approx$ 1.773? $\endgroup$ – Cozen Oct 29 '13 at 0:15
  • $\begingroup$ So there was an 88% error? Lol I don't think thats correct $\endgroup$ – Cozen Oct 29 '13 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.