Answer
$f(x) \approx 1-\frac{x}{2}$
$\sqrt{0.9} \approx 0.95$
$\sqrt{0.99} \approx 0.995$
We can see a sketch of the graphs below.
Work Step by Step
$f(x) = \sqrt{1-x}$
$f'(x) = -\frac{1}{2~\sqrt{1-x}}$
When $x = 0$:
$f'(x) = -\frac{1}{2~\sqrt{1-x}}$
$f'(0) = -\frac{1}{2~\sqrt{1-0}}$
$f'(0) = -\frac{1}{2}$
We can find the linear approximation at $a=0$:
$f(x) \approx f(a)+f'(a)(x-a)$
$f(x) \approx f(0)+f'(0)(x-0)$
$f(x) \approx 1+(-\frac{1}{2})(x)$
$f(x) \approx 1-\frac{x}{2}$
We can approximate $\sqrt{0.9} = f(0.1)$:
$f(x) \approx 1-\frac{x}{2}$
$f(0.1) \approx 1-\frac{0.1}{2}$
$f(0.1) \approx 0.95$
$\sqrt{0.9} \approx 0.95$
We can approximate $\sqrt{0.99} = f(0.01)$:
$f(x) \approx 1-\frac{x}{2}$
$f(0.01) \approx 1-\frac{0.01}{2}$
$f(0.01) \approx 0.995$
$\sqrt{0.99} \approx 0.995$
We can see a sketch of the graphs below.