Question

Find the equation of the tangent line of the given function at the given point.
g(x) =1/square root x , at (1,1)

Answer 1

\[g(x)= 1/\sqrt{x}\]
\[g'(x)= -1/2\sqrt{x}\]
slope of tangent at 1,1 will be -1/2
so the equation of tangent becomes
(y-1)= -1/2 (x-1)
x+2y=3 (final equation)