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

No positive value of x makes the tangent line intersect the y-axis at (0,2).

Step-by-step explanation:

To find the positive value of x that makes the tangent line to the curve y=ln(1-x) intersect the y-axis at the point (0,2), we need to find the x-coordinate of the point where the tangent line intersects the y-axis, which is when x=0. We can substitute x=0 into the equation y=ln(1-x) to find the y-coordinate of the point, which is ln(1-0)=ln(1)=0. So, the point where the tangent line intersects the y-axis is (0,0).

Since we are given that the point of intersection is (0,2), this means that the line passes through the point (0,2) and (0,0). We can find the slope of the line using the formula m = (y2-y1)/(x2-x1), where (x1,y1) and (x2,y2) are the coordinates of the given points.

Substituting (0,2) and (0,0), we get m = (0-2)/(0-0) = -2/0. Since the denominator is zero, the slope is undefined.

Therefore, there is no positive value of x that makes the tangent line to the curve y=ln(1-x) intersect the y-axis at the point (0,2).

Answer 2

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