Question

Find the equations of the tangent line to curve y=(x-1)/(x 1).

Answer 1

The equation of the tangent line to the curve y = (x-1)/(x+1) is found by calculating the derivative to determine the slope, and then using the point-slope form of the equation for a specific value of x.

Step-by-step explanation:

To find the equation of the tangent line to the curve y = (x-1)/(x+1), we first need to determine the slope of the tangent line at a particular point. This is done by calculating the derivative of the given function. Once we have the derivative, we can find the slope of the tangent line for a specific value of x by plugging that value into the derivative. The equation of the tangent line can then be found using the point-slope form.

Let's find the derivative of y = (x-1)/(x+1). Using the quotient rule, the derivative y' is ((x+1)(1) - (x-1)(1))/((x+1)^2) = (2)/(x+1)^2. Now, let's assume we want to find the tangent line at x = a. The slope at this point is (2)/(a+1)^2. If y(a) is the y-coordinate of the point where the tangent line touches the curve, the equation of the tangent line is y - y(a) = ((2)/(a+1)^2)(x - a).

Without a specific point, we cannot find the exact equation of the tangent line. If you provide a specific value for x, we can plug it into the derivative to find the slope at that point and then determine the exact equation of the tangent line.

Answer 2

The equations of the tangent lines to y = (x-1)/(x+1) that are parallel to x - 2y = 5 are:

2y = x - 1

2y = x + 7

Step-by-step explanation:

Parallel Lines: Parallel lines have equal slopes. To find the slope of the tangent lines, we need to differentiate the given curve y = (x-1)/(x+1).

Differentiating the Curve: Using the quotient rule, we get:

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

Matching Slopes: Since the parallel lines share the same slope as x - 2y = 5, we need to equate their slopes. Re-arranging x - 2y = 5 to isolate y, we get:

2y = x - 5

Therefore, the slope of both tangent lines is 2.

Specific Tangent Lines: With the common slope of 2, we need to find the specific x-coordinates where the tangent lines touch the curve. These points will serve as the y-intercepts for the tangent line equations.

Tangential Point 1: Let's assume one tangent line touches the curve at x = a. Substituting this into y = (x-1)/(x+1) and equating it to 2 (slope), we get:

2 = (a-1)/(a+1)

Solving for a, we find a = 3. This gives us the first tangent point at (3, 2/4) = (3, 1/2).

Tangential Point 2: Similarly, assume another tangent line touches the curve at x = b. Solving the same equation as before, we get b = -7. This gives us the second tangent point at (-7, -2/4) = (-7, -1/2).

Tangent Line Equations: Finally, using the point-slope form of linear equations with the respective slopes and points, we get the tangent line equations:

2y = x - 1 (passes through (3, 1/2))

2y = x + 7 (passes through (-7, -1/2))

Therefore, these two equations represent the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x - 2y = 5.