Question
the tangent to the curve y=a/x +bx at (1,3) is parallel to the line whose equation is y=2x+1. find the values of a and b.
Answers
Reiny
First of all, the point (1,3) is on the curve, so
3 = a/1 + b(1)
a+b = 3
dy/dx = -a/x^2 + b
at (1,3) , dy/dx = -a + b
but it is parallel to y = 2x+1, so the slope is 2
-a + b = 2
add our two equations in a and b:
2b = 5
b = 5/2
sub back into a+b=3
a+5/2 = 3
a = 1/2
a = 1/2, b = 5/2
3 = a/1 + b(1)
a+b = 3
dy/dx = -a/x^2 + b
at (1,3) , dy/dx = -a + b
but it is parallel to y = 2x+1, so the slope is 2
-a + b = 2
add our two equations in a and b:
2b = 5
b = 5/2
sub back into a+b=3
a+5/2 = 3
a = 1/2
a = 1/2, b = 5/2