Respuesta :
Answer:
0 < x < 1
Step-by-step explanation:
the derivative of the curve allows one to compute the slope:
[tex]y=ln(1-x)\\\\\frac{dy}{dx}=\frac{-1}{1-x}=\frac{1}{1-x}[/tex]
(1/1-x) is the value of the slope, which depends of the point x. By using the line equation we have:
[tex]y_l=mx_l+b=(\frac{1}{x-1})x+b[/tex]
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:
[tex]y_l=\frac{x}{x-1}+2[/tex]
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!!
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
