calculus
posted by Anonymous .
Determine the coordinates of the point on the graph of f(x)= (2x+1)^(1/2) where the tangent line is perpendicular to the line 3x+y+4=0

dy/dx = (1/2) (2x+1)^(1/2) (2)
= 1/(2x+1)^(1/2) or 1/√(2x+1)
slope of line is 3
so slope of perpendicular is 1/3
so we have ...
1/√(2x+1) = 1/3
√(2x+1) = 3
square both sides ...
2x + 1 = 9
2x = 8
x = 4
f(4) = (9)^1/2) = 3
so the point of contact is (4,3)
and the equation of the tangent is
y3 = (1/3)(x4)
3y  9 = x4
x  3y = 5 OR y = (1/3)x + 5/3
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