# Am I Doing a Linear Approximation on $\sqrt x$ Correctly?

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?

• 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. – 0XLR Jul 29 at 4:58
• Looks good to me. – Andrei Jul 29 at 4:58
• 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). – runway44 Jul 29 at 4:58
• Thanks. I was wondering how people do that. – Burt Jul 29 at 4:59
• 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. – 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.