Estimate Δf using the Linear Approximation and use a calculator to compute the error. $f(x)= \sqrt{3+x}$


$Δx= 0.5$

$Δf \approx \frac{1}{12}$

^ I got this part, just take the derivative.

With these calculations, we have determined that the square root of _____

^I was thinking it was the $\sqrt9$, but that is incorrect

is approximately _____

^obviously thought this was $3$

The error in Linear Approximation is: _____

^Went through the whole process of trying to find the approximation, ended up with a $negative$ number though $-107.38\%$

[Note: This is not asking for relative error or percent error.]

My main struggle is trying to understand what they are asking for? Any ideas?

  • $\begingroup$ I've no idea what your first few lines mean. $\endgroup$ – user21820 Mar 30 '16 at 4:27
  • $\begingroup$ I was hoping I wasn't the only one. $\endgroup$ – pewpew Mar 30 '16 at 4:28
  • $\begingroup$ Are you sure the question states exactly that? What is $a$? $\endgroup$ – user21820 Mar 30 '16 at 4:31
  • $\begingroup$ Maybe you're supposed to find $\sqrt{9.5} using linear approximation and then determine the error. $\endgroup$ – Neal Mar 30 '16 at 4:32
  • $\begingroup$ @Neal: That's what I would guess also. $\endgroup$ – user21820 Mar 30 '16 at 4:33

$$\quad{f(X+\Delta x) \sim f(x) +\Delta x .f'(x)\\f(x)=\sqrt{3+x},f'(x)=\frac{1}{2\sqrt{3+x}}\\\to\\f(6+.5)\sim f(6) +.5 \frac{1}{2\sqrt{9}}=3+0.5*\frac{1}{6}=3+\frac{1}{12}$$ now $$f(6+\Delta x)-f(6)\sim \frac{1}{12}=0.083333333...}$$

without approximation $$\sqrt{3+6+0.5}-\sqrt{3+6}=0.0822070014844...$$


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