Linear approximation to $y = \sqrt{1-x}$ at $x=0$, then approximate $\sqrt{0.9}$ and $\sqrt{0.99}$

Question

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?

Answer 1

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.

Answer 2

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+$\frac{-1}{2}$ $x$

Answer 3

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 $1-x$.