Question

For what positive value of x does the tangent line to the curve y=ln(1-x) intersect the y-axis at the point (0,2)

Answer 1

Answer:

x = 1/2

Step-by-step explanation:

Solution:-

- Any line tangent to a curve defined by y = f(x) have the gradient -slope equal to the first derivative of the curve, dy/dx  :

                                 y = Ln ( 1 - x )

                                 dy/dx = -1 / ( 1 - x )

- So the line tangent to the curve has slope defined by the first derivative evaluated, while the equation of tangent line is:

                                 y = mx + c

                                 m = dy/dx = -1 / ( 1 - x )

Where,                      c = y - intercept = 2

                                 y = -x / ( 1 - x ) + 2 ..... Equation of tangent

- The x-point of intersection can be evaluated by simultaneously solving the equation of tangent and curve y = f(x):

                                 -x + 2 - 2x = ( 1 - x )*Ln ( 1 -x )

                                 -3x + 2 = Ln ( 1 -x ) - Ln ( 1 - x )^x

                                 -3x + 2 = Ln ( 1 - x ) ^ ( 1 - x )

                                 ( 1 - x ) ^ ( 1 - x ) = e^ ( -3x + 2 )

                                  1 - x = e , x = 1-e

                                  1 - x = -3x + 2, x = 1/2

- The x-value of point of intersection is x = 1/2

Answer 2

Answer:

0 < x < 1

Step-by-step explanation:

the derivative of the curve allows one to compute the slope:

$y=ln(1-x)\\\\\frac{dy}{dx}=\frac{-1}{1-x}=\frac{1}{1-x}$

(1/1-x) is the value of the slope, which depends of the point x. By using the line equation we have:

$y_l=mx_l+b=(\frac{1}{x-1})x+b$

where yl is the tangent line of the original curve y in the point x.

For the intersection point (0,2), with b=2, we obtain:

$y_l=\frac{x}{x-1}+2$

The denominator must be different of zero. And the original curve y(x) only can take values of x < 1. Hence the positive value of x for the tangent line will be:

0 < x < 1

hope this helps!!