Find the tangent line approximation for sqrt(3+x) near x=0.

Answer 1

when x = 0, y = sqrt 3

y = (x+3)^.5
so
dy/dx = .5 (x+3)^-.5
slope of tangent line at x = 0 is thus
.5/sqrt 3
y = m x + b
sqrt 3 = (.5/sqrt 3)(0) + b
b = sqrt 3

y = (.5/sqrt 3) x + sqrt 3

Answer 2

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