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56. How many tangent lines to the curve
y=x /(x+1) pass through the point (1,2)? At which points do these tangent lines touch the curve?

User Nile
by
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1 Answer

12 votes

There are 2 tangent lines that pass through the point


y=(1)/((-1+√(3)^2) ) (x-1)+2

and


y=(1)/((-1-√(3)^2) ) (x-1)+2

Step-by-step explanation:

Given:


y=(x)/(x+1)

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:


y=m(x-1)+2
[1]

For the lines to be tangent to the curve, we must substitute the first derivative of the curve for
m:


(dy)/(dx) =(d(x))/(dx)(x+1)-x^(d(x+1))/(dx) \\ \\


(dy)/(dx) =(x+1-x)/((x+1)^2)


(dy)/(dx)= (1)/((x+1)^2)


m=(1)/((x+1)^2)
[2]

Substitute equation [2] into equation [1]:


y=(x-1)/((x+1)^2)+2
[1.1]

Because the line must touch the curve, we may substitute
y=(x)/(x+1):


(x)/(x+1)=(x-1)/((x+1)^2)+2

Solve for x:


x(x+1)=(x-1)+2(x+1)^2


x^2+x=x-1+2x^2+4x+2


x^2+4x+1


x(-4±√(4^2-4(1)(1)) )/(2(1))


x=-2 ±
√(3)


x=-2 ±
√(3)
and
x=-2-√(3)

There are 2 tangent lines.


y=(1)/((-1+√(3)^2) ) (x-1)+2

and


y=(1)/((-1-√(3)^2) ) (x-1)+2

User Spaceman Spiff
by
6.8k points