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The given function was

$$f(x)=ln(\frac{2}{x})$$

and I had to compute the linear approximation at x = 2. I obtained the answer of

$$L(x)=-\frac{1}{2}(x-2)$$

I am then supposed to use that approximation to estimate $ln(1.9)$.

At this point I don't understand what to do, because my class only covered examples where the x had the same modifier, i.e. the function was $f(x)=\sqrt{x}$ and the point we had to estimate was $\sqrt2$, and a search on the web only returns the same types of examples.

I assume I have to use a number that will be nicely divisible by 2 and close to 1.9, so I would choose 2. Therefore, $\frac{2}{2}=1$, and when I plug that in to $L(x)$ I would obtain .5, which is a slight underestimate, and

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

so the concavity confirms that it is an underestimate.

I think this is the solution, but I just had to make sure. Is my methodology correct? Should I have used a different number for the estimate, should I have changed the number at all? Any help will be great. Thank you!

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  • $\begingroup$ If you have the linear approximation at x=2, just plug in 1.9 into $L(x)$. The approximation you have is accurate for any value "near" 2 and 1.9 is pretty close to 2. $\endgroup$ – JessicaK Nov 10 '14 at 1:38
  • $\begingroup$ That's what I initially thought too! But, it gives a completely inaccurate answer (.05), so I was led to think that it was a problem with the modifier of 2/x. Since the number I plug in would essentially just be x, I assumed that you had to modify the number to be easy to work with and be a form of 2/x $\endgroup$ – mrybak834 Nov 10 '14 at 1:44
  • $\begingroup$ $\ln(2/1.9) = 0.051293294$, and $L(1.9) = 0.05$. Accuracy up to 2 decimal places seems pretty good to me for an approximation you can do on a napkin. If you wanted more accuracy, you would center the approximation closer to 1.9 or use more derivatives which you will presumably see later when you cover Taylor's Theorem. $\endgroup$ – JessicaK Nov 10 '14 at 1:47
  • $\begingroup$ Not exactly. You are right that L(1.9) = .05, and that's the issue. The actual result of ln(1.9) = .6. Considering the answer I obtained was .5, not .05, this is a huge difference and makes me think all the more that my answer was correct due to how close it is. $\endgroup$ – mrybak834 Nov 10 '14 at 1:52
  • $\begingroup$ You defined $f(x):= \ln(2/x)$. Why are you calculating $\ln(1.9)$? $\endgroup$ – JessicaK Nov 10 '14 at 1:53

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