Calculus!!??

Question

Find two equations of the tangent lines to the curve below that are parallel to the line x - 2y = 2.

y = (x-1)/(x+1)

Answer 1

The line has gradient 1/2

y = (x-1)/(x+1) = 1 - 2/(x+1)
dy/dx = 2/(x+1)^2 = 1/2
(x+1)^2 = 4
x+1 = +/- 2
x = -3, 1
y = 2, 0

The respective equations are:
x - 2y = -7 and
x - 2y = 1

Answer 2

The tangent line is supposed to be parallel to the line x - 2y = 2 (which can be re-written as y = 1/2 x - 1). Thus the tangent line has to have slope of of 1/2.

Determine where the curve y = (x-1) / (x+1) has tangent lines with slopes 1/2. To do this set the first derivative equal to 1/2 (you'll get 2 solutions x = -3, 1 ).

Each solution will give you one tangent line.

Take for example the solution x = 1.
First determine the value of y on the CURVE when x = 1 ( y = 0 )

Using the point slope form of a line we can write down the answer.

( y - 0 ) / ( x - 1 ) = 1 / 2.

That's the equation of 1 of the tangent lines. The answer in the back of the book might look different.

Good Luck!!!

Answer 3

There are a few steps involved.

1) Find the derivative of the curve, say it's y = f(x)
f'(x) = 2/((x+1)^2) ... I assume you know how to do this.

2) Find the slope of the line.
Rewriting the line as y = x/2 - 1 shows slope is 1/2.

3) Find the x-values of the points for which the slope is the same as the slope of the line.
2/((x+1)^2) = 1/2 gives (x+1)^2 = 4, so x+1 = 2, -2 therefore x = 1, -3 are the two x-values.

4) Find the tangent points.
Just plug the x-values you got into the function: f(1) = 0, f(-3) = 2 and so the points are (1,0) and (-3,2).

5) Find the tangent lines.
These are the lines for which slope is 1/2 and passing through (1,0) and (-3,2) respectively. You know how to do this!

After you're done, make sure and graph all 3 functions to be certain that your results are correct!

Answer 4

y = (x-1)/(x+1)
y' = [(x+1)-(x-1)]/(x+1)^2 = 2/(x+1)^2
Since x-2y =2 has a slope of 1/2, the slope of the lines parallel to it must also have a slope of 1/2
So 2/(x+1)^2 = 1/2
(x+1)^2 = 4
x+1 = +/- 2
x = - 1 +/- 2
x = -3 or 1
When x = -3,f(x) = 2
So y=x/2 +b
2 = -3/2 +b --> b = 3.5
So y= x/2 + 3.5 is one tangent line
When x = 1, f(x) = 0
0 = 1/2 =b --> b = -.5
So y=x/2 -.5 is the other tangent line

Answer 5

y=(1/2)x-1

slope=1/2

y=(x-1) (x+1)**(-1)
deriv(y)=(1/(x+1)) - (x-1)/((x+1)**-2)
=2/(x+1)**2)
set deriv(y)=0.5
4=(x+1)**2

x+1=2 or x+1=-2

x=1 or x=-3

(1,0) and (-3,2) are where tangents are and have slope=0.5

for (1,0) ===> y-0=0.5(x-1)

y=0.5 x - 0.5 (answer)


for (-3,2) ===> y-2 = 0.5 (x+3)

y=0.5 x + 3.5 (answer)

Answer 6

x-2y=2