Given the function $y = x^3 - x$ with graph C. a. Show that A(-1, 0) is an intersection point between the graph C and the line L with the equation $y = a(x+1)$. b. Find the equation of the tangent line to graph C at point A. c. Find the value of $a$ such that $L$ intersects graph $C$ at two other distinct points $M_1$ and $M_2$ besides the point $A$. Find the set of midpoints $I$ of the segment $[M_1M_2]$.

1. Problem Description

Given the function $y = x^3 - x$ with graph C.

a. Show that A(-1, 0) is an intersection point between the graph C and the line L with the equation $y = a(x+1)$.

b. Find the equation of the tangent line to graph C at point A.

c. Find the value of $a$ such that $L$ intersects graph $C$ at two other distinct points $M_1$ and $M_2$ besides the point $A$. Find the set of midpoints $I$ of the segment $[M_1M_2]$.

2. Solution Steps

a. To show that $A(-1, 0)$ is an intersection point between the graph $C: y = x^3 - x$ and the line $L: y = a(x+1)$, we substitute $x = -1$ and $y = 0$ into both equations.

For the graph C:

$y = x^3 - x$

$0 = (-1)^3 - (-1)$

$0 = -1 + 1$

$0 = 0$

So, point $A(-1, 0)$ lies on the graph $C$.

For the line L:

$y = a(x+1)$

$0 = a(-1+1)$

$0 = a(0)$

$0 = 0$

So, point $A(-1, 0)$ lies on the line $L$.

Therefore, $A(-1, 0)$ is an intersection point between graph $C$ and line $L$.

b. To find the equation of the tangent line to graph $C$ at point $A(-1, 0)$, we first find the derivative of the function $y = x^3 - x$.

$\frac{dy}{dx} = 3x^2 - 1$

The slope of the tangent line at $x = -1$ is:

$m = 3(-1)^2 - 1 = 3(1) - 1 = 3 - 1 = 2$

The equation of the tangent line at point $A(-1, 0)$ is given by:

$y - y_1 = m(x - x_1)$

$y - 0 = 2(x - (-1))$

$y = 2(x + 1)$

$y = 2x + 2$

c. To find the value of $a$ such that the line $L: y = a(x+1)$ intersects the graph $C: y = x^3 - x$ at two other distinct points besides $A(-1, 0)$, we set the equations equal to each other:

$x^3 - x = a(x+1)$

$x^3 - x - a(x+1) = 0$

$x^3 - x - ax - a = 0$

$x^3 - a - x - ax = 0$

$x^3 - x - a(x + 1) = 0$

$(x+1)(x^2 - x - a) = 0$

Since $x = -1$ corresponds to point A, we are interested in the solutions of the quadratic equation $x^2 - x - a = 0$. For the line to intersect the curve at two other distinct points $M_1$ and $M_2$, this quadratic equation must have two distinct real roots.

The discriminant of the quadratic equation is:

$\Delta = (-1)^2 - 4(1)(-a) = 1 + 4a$

For two distinct real roots, $\Delta > 0$.

$1 + 4a > 0$

$4a > -1$

$a > -\frac{1}{4}$

Let $x_1$ and $x_2$ be the roots of the equation $x^2 - x - a = 0$. These are the x-coordinates of the intersection points $M_1$ and $M_2$.

By Vieta's formulas, $x_1 + x_2 = 1$.

Let $I(x, y)$ be the midpoint of the segment $[M_1M_2]$. Then,

$x = \frac{x_1 + x_2}{2} = \frac{1}{2}$

$y = a(x+1) = a(\frac{1}{2} + 1) = a(\frac{3}{2}) = \frac{3a}{2}$

Since $a > -\frac{1}{4}$, we have $y > \frac{3}{2}(-\frac{1}{4}) = -\frac{3}{8}$

Therefore, the midpoint $I$ has coordinates $(\frac{1}{2}, \frac{3a}{2})$, where $a > -\frac{1}{4}$.

The set of midpoints is the line $x = \frac{1}{2}$, and $y > -\frac{3}{8}$.

3. Final Answer

a. Shown that $A(-1, 0)$ is an intersection point.

b. The equation of the tangent line is $y = 2x + 2$.

c. The value of $a$ is $a > -\frac{1}{4}$. The set of midpoints $I$ is the line $x = \frac{1}{2}$, and $y > -\frac{3}{8}$.