This was the question assigned to me:

Let $f$ be the function defined by $f(x)=\sqrt x$. Using the line tangent to the graph at $x=9$, what is the approximation of $f(9.3)$?

I know the formula for linear approximation is $$f(x)+f'(x)\Delta x$$ I plugged it in and got: $$3+\frac16(.3)$$ Doing the simple calculation results in $3.05$.
I do not see any problems with how I did this problem. However, I am a beginner and feel I may be missing something. Was this done correctly?

  • 2
    $\begingroup$ Yes that is correct. Your steps are correct. If you are still unsure about your anwser, you can check that by calculating $\sqrt{9.3}$ on google which is $\approx 3.0496$ and that is pretty close to your answer. $\endgroup$ – 0XLR Jul 29 at 4:58
  • $\begingroup$ Looks good to me. $\endgroup$ – Andrei Jul 29 at 4:58
  • 1
    $\begingroup$ Looks right to me. Also you can do blockquotes by putting > in front of text, and the preview function should indicate when your post is unreadable (e.g. when latex goes off-screen). $\endgroup$ – runway44 Jul 29 at 4:58
  • $\begingroup$ Thanks. I was wondering how people do that. $\endgroup$ – Burt Jul 29 at 4:59
  • $\begingroup$ Just a very minor point : write $f(x+\Delta x)=f(x)+f'(x)\Delta x$ or, better $f(x+\Delta x)\approx f(x)+f'(x)\Delta x$. Otherwise, this is very correct. $\endgroup$ – Claude Leibovici Jul 29 at 6:49

This looks correct. In fact, if you use a calculator to compute $\sqrt{9.3},$ you’ll see that your linear approximation is very good.


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