0
$\begingroup$

How do I find this? I know that the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$. Here, I would plug in $(1-x)$ instead of $x$. When $x = 0$, the slope would evaluate to $\dfrac{1}{2}$. I got that the approximation near $0$ would be $\frac{x}{2}+\frac{1}{2}$. The approximation was correct for $\sqrt{0.9}$ and $\sqrt{0.99}$, but not for $\sqrt{1}$. I tried to submit the approximation for $\sqrt{1}$ as $1$, but that was incorrect. Therefore, the approximation does not work for $x = 0$. What do I do next? Have I made a mistake somewhere in the process?

$\endgroup$
  • $\begingroup$ Your derivative is incorrect. You cannot just "plug in" $1-x$, you must use the chain rule. $\endgroup$ – George V. Williams Apr 13 '13 at 19:03
2
$\begingroup$

Let $f(x) = \sqrt{1-x}$, then use the chain rule with $u=1-x$ (this is where your mistake was), and get that $f'(x) = -\frac{1}{2\sqrt{1-x}}$. $f(0) = 1$, and $f'(0) = -\frac12$. So our approximation is:

$$ f(x) \approx 1 - \frac x 2$$

Now just plug in the required values.

$\endgroup$
  • $\begingroup$ Rather $f(x)\approx 1-\frac12x$. $\endgroup$ – Hagen von Eitzen Apr 13 '13 at 19:12
  • $\begingroup$ @HagenvonEitzen, thanks. $\endgroup$ – George V. Williams Apr 13 '13 at 19:14
  • $\begingroup$ Thanks guys. At least I didn't make errors. Turned out that the answer box in webwork expected the equation, not its value. $\endgroup$ – cuabanana Apr 13 '13 at 19:25
2
$\begingroup$

The correct approximation is $y\approx f(0)+xf'(0)=1-\frac12x$. Sign errors can happen, but I have no idea where you got the constant part of $\frac12$ from. Thus you should obtain $\sqrt{0.9}\approx 0.95$ and $\sqrt{0.99}\approx 0.995$ and of course $\sqrt 1"\approx" 1$.

I guess you messed up $x$ vs. $1-x$ at least once because if your approximation is good enough to find good approximations for $\sqrt{0.9}$ (i.e.$x=0.1$) and $\sqrt{0.99}$ (i.e. $x=0.01$) then it will also work for $\sqrt 1$ (i.e. $x=0$). Make sure you always are clear about when to use $x$ and when $1-x$.

$\endgroup$
1
$\begingroup$

You need to get the derivative of $y=\sqrt(1-x)$, which is -1/2$\sqrt(1-x)$. Now will get the correct results: $\sqrt(1-x)$=$f(x_0)$+$\frac{-1}{2\sqrt(1-x_0)}$ $\Delta x$. If $x_0=0$, then $\sqrt(1-x)$=1+$\frac{-1}{2}$ $x$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.