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erik [133]
2 years ago
11

56. How many tangent lines to the curve

x+1)" alt="y=x /(x+1)" align="absmiddle" class="latex-formula"> pass through the point (1,2)? At which points do these tangent lines touch the curve?
Mathematics
1 answer:
PIT_PIT [208]2 years ago
8 0

There are 2 tangent lines that pass through the point

y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2

and

y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2

Explanation:

Given:

y=\frac{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:

\frac{dy}{dx} =\frac{d(x)}{dx}(x+1)-x^\frac{d(x+1)}{dx} \\ \\

\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}

\frac{dy}{dx}= \frac{1}{(x+1)^2}

m=\frac{1}{(x+1)^2} [2]

Substitute equation [2] into equation [1]:

y=\frac{x-1}{(x+1)^2}+2 [1.1]

Because the line must touch the curve, we may substitute y=\frac{x}{x+1}:

\frac{x}{x+1}=\frac{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\frac{-4±\sqrt{4^2-4(1)(1)} }{2(1)}

x=-2 ± \sqrt{3}

x=-2 ± \sqrt{3}<em> </em>and x=-2-\sqrt{3}

There are 2 tangent lines.

y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2

and

y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2

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