Find the linear approximation of the tangent line to f(x)=1/the square root of 1-x at x=0

Question

anonymous · Answer

-1/2 i think

anonymous · Answer

f(x) = 1/sqrt(1-x)

f'(x) = -1/2sqrt(1-x)

f'(0) = -1/2

anonymous · Answer

Linear approximation around x = x0 in general:
$$y - f(x_{0}) = f'(x_{0})(x - x_{0})$$
$$y= f'(x_{0})(x - x_{0}) +f(x_{0})$$

$$f(x) = 1/\sqrt{1-x}$$
$$f'(x) = 1/(2\sqrt{(1 - x)^3})$$
so our approximation:
$$f(0) = 1$$
$$f'(0) = 1/2$$
$$y = (1/2)(x - 0) + 1$$
$$y = (1/2)x + 1$$

anonymous · Answer

oh man...i got that basic derivative wrong...sorry...ignore my nonsensical post