How many tangent lines to the curve y=xx+1 pass through the point (1,2)

The form of the tangent lines must be: y=m(x−1)+2 [1] Now, substitute the first derivative of the curve for m dydx=d(x)dx(x+1)−xd(x+1)dx(x+1)2 dydx=x+1−x(x+1)2 dydx=1(x+1)2 m=1(x+1)2 [2] Now, substitute equation [2] into equation [1] y=x−1(x+1)2+2 [1.1] We have to substitute y=xx+1 xx+1=x−1(x+1)2+2 Solve for x x(x+1)=(x−1)+2(x+1)2 x2+x=x−1+2x2+4x+2 x2+4x+1 x=−4±42−4(1)(1)2(1) x=−2±3 x=−2+3 and x=−2−3 There are two tangent lines y=1(−1+3)2(x−1)+2 and y=1(−1−3)2(x−1)+2

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Joan Thompson

Answered question

2021-12-13

How many tangent lines to the curve

y = \frac{x}{x + 1}

pass through the point (1,2)

Answer & Explanation

Ana Robertson

Beginner2021-12-14Added 26 answers

The form of the tangent lines must be:

y = m (x - 1) + 2

[1]

Now, substitute the first derivative of the curve for m

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

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

\frac{d y}{d x} = \frac{1}{{(x + 1)}^{2}}

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

[2]

Now, substitute equation

[2]

into equation

[1]

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

[1.1]

We have to 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 + 2 x^{2} + 4 x + 2

x^{2} + 4 x + 1

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

x = - 2 \pm \sqrt{3}

x = - 2 + \sqrt{3}

and

x = - 2 - \sqrt{3}

There are two 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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