Using $f(x, y, z) = \sqrt{x + 2y + 3z}$ and an appropriately chosen linear approximation to find the value of $\sqrt{226}$
If I take $ y= \sqrt{x}$ centered at 0 and expand the series approximation, unsure how to use the given function in any place. A
How to proceed with this? Any hint would be appreciated.
Consider the special case $x\gg y,\,z$ so$$f=\sqrt{x}\sqrt{1+\frac{2y+3z}{x}}\approx\sqrt{x}\left(1+\frac{2y+3z}{2x}\right)=\sqrt{x}+\frac{2y+3z}{2\sqrt{x}}.$$Now choose $x=15^2,\,y=\frac12,\,z=0$ to get $\sqrt{226}\approx15+\frac{1}{30}$. To $6$ decimalplaces this is $15.033333$, whereas the correct answer is $15.033296$.