Question

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

Answer 1

-1/2 i think

Answer 2

f(x) = 1/sqrt(1-x)

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

f'(0) = -1/2

Answer 3

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:
\[f(0) = 1\]
\[f'(0) = 1/2\]
\[y = (1/2)(x - 0) + 1\]
\[y = (1/2)x + 1\]

Answer 4

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