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

1. Problem Description

The problem is about the function $y = x^3 - x$ and its graph $C$.

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

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

c) Find the value of $a$ such that $L$ intersects the graph $C$ at two distinct points $M_1$ and $M_2$ other than $A$. 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 of the graph $C$ and the line $L$, we need to verify that the coordinates of $A$ satisfy both equations.

For the graph $C$, $y = x^3 - x$. Substituting $x = -1$, we get $y = (-1)^3 - (-1) = -1 + 1 = 0$. So, the point $A(-1, 0)$ lies on the graph $C$.

For the line $L$, $y = a(x+1)$. Substituting $x = -1$, we get $y = a(-1+1) = a(0) = 0$. So, the point $A(-1, 0)$ lies on the line $L$.

Thus, $A(-1, 0)$ is an intersection point of the graph $C$ and the line $L$.

b) To find the equation of the tangent line to the graph $C$ at point $A$, we need to find the derivative of $y = x^3 - x$ with respect to $x$.

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

At point $A(-1, 0)$, the slope of the tangent line 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

$y - y_1 = m(x - x_1)$, where $(x_1, y_1) = (-1, 0)$.

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

$y = 2(x + 1)$

$y = 2x + 2$

The equation of the tangent line to the graph $C$ at point $A$ is $y = 2x + 2$.

c) We need to find the value of $a$ such that $L$ intersects the graph $C$ at two distinct points $M_1$ and $M_2$ other than $A$.

We have $y = x^3 - x$ and $y = a(x+1)$.

Equating the two equations:

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

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

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

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

$(x+1)[x(x-1) - a] = 0$

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

One solution is $x = -1$, which corresponds to point $A$.

The other two solutions come from the quadratic equation $x^2 - x - a = 0$.

For $L$ to intersect $C$ at two distinct points other than $A$, the quadratic equation must have two distinct real roots.

The discriminant of the quadratic equation is $\Delta = b^2 - 4ac = (-1)^2 - 4(1)(-a) = 1 + 4a$.

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

$1 + 4a > 0$

$4a > -1$

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

Also, we need to ensure that $x \neq -1$. Substituting $x = -1$ into the quadratic:

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

$1 + 1 - a = 0$

$2 - a = 0$

$a = 2$

So, we need $a > -\frac{1}{4}$ and $a \neq 2$.

Now we find the midpoint $I$ of the segment $[M_1 M_2]$.

Let $x_1$ and $x_2$ be the two distinct roots of the equation $x^2 - x - a = 0$.

The x-coordinate of the midpoint $I$ is $x_I = \frac{x_1 + x_2}{2}$.

From Vieta's formulas, we know that the sum of the roots of the quadratic equation $x^2 - x - a = 0$ is $x_1 + x_2 = -\frac{-1}{1} = 1$.

So, $x_I = \frac{1}{2}$.

The y-coordinate of the midpoint $I$ is $y_I = a(x_I + 1) = a(\frac{1}{2} + 1) = a(\frac{3}{2}) = \frac{3}{2}a$.

Since $a > -\frac{1}{4}$ and $a \neq 2$, we have $y_I > -\frac{3}{8}$ and $y_I \neq 3$.

So, the set of midpoints $I$ is the horizontal line $x = \frac{1}{2}$, with $y > -\frac{3}{8}$ and $y \neq 3$.

3. Final Answer

a) $A(-1, 0)$ is an intersection point of the graph $C$ and the line $L$.

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

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