SOLUTION: (a) Find the value of m for which the line y=mx-3 is a tangent to the curve y=x+1/x and find the x-coordinate of the point at which this tangent touches the curve. (b) Find the

Question 463056: (a) Find the value of m for which the line y=mx-3 is a tangent to the curve y=x+1/x and find the x-coordinate of the point at which this tangent touches the curve.
(b) Find the values of c and d for which {x:-5 ps. I have the answers but i do not know how to derive to the answer. Your help will be appreciated. Thank you:)
answers: (a)m=-5/4, x=-2/3 (b)c=2, d=15
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
(a) Find the value of m for which the line y=mx-3 is a tangent to the curve y=x+1/x and find the x-coordinate of the point at which this tangent touches the curve.
(b) Find the values of c and d for which {x:-5 ps. I have the answers but i do not know how to derive to the answer. Your help will be appreciated. Thank you:)
answers: (a)m=-5/4, x=-2/3 (b)c=2, d=15
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(a) The derivative of the curve at a point (x0,y0) is the slope of the tangent line at that point.
The curve and its tangent line intersect at the point (x0,y0).
The equation for the curve is y = x + 1/x
The equation for the tangent line is y = mx - 3
y = x + 1/x
y = mx - 3
So we can equate the RHS's:
x + 1/x = mx - 3 [1]
But the slope of the tangent line, dy/dx = m = 1 - 1/x^2 [2]
Solve for m in [1] and substitute into [2]
x + 1/x + 3 = mx
1 + 1/x^2 + 3/x = m
1 + 1/x^2 + 3/x = 1 - 1/x^2
2/x^2 + 3/x = 0
2 + 3x = 0
So x0 = -2/3
Since the point (x0,y0) lies on the curve y = x + 1/x, we have y0 = -2/3 + 1/(-2/3)
This gives y0 = -13/6
So the slope of the tangent line is (y0 - b)/(x0 - 0) = (-13/6 + 3)/(-3/2) = -5/4
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