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posted by Mike Sunday, March 3, 2013 at 4:35pm.

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

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

Thank you

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