my derivatiave is (1/2)(2x+1)^(-1/2)(2)
= 1/√(2x+1) , that is the slope of the tangent.
slope of 3x+y+4 = -3
so the slope of the perpendicular is 1/3
then 1/√(2x+1) = 1/3
square both sides and crossmultiply
2x+1 = 9
x = 4
sub x=4 into original to finish
Determine the coordinates of the point on the graph of f(x)=sqrt(2x+1) where the tangent line is perpendicular to the line 3x+y+4=0
I do not seem to get the answer... I find the first derivative of f(x0 and equate it to the value of the slope from the tangent line. the answer says it should be (4,3) but i do not even get the x-value correct.
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