The problem states that we have a function $f_a(x) = x^3 - 1 - a(x-1)$ and its graph $C_a$ depending on the parameter $a$. We need to: a) Find the value(s) of $a$ such that the graph $C_a$ is tangent to the x-axis. b) Study the variation and sketch the graph of the function for the value(s) of $a$ found in part (a).

1. Problem Description

The problem states that we have a function $f_a(x) = x^3 - 1 - a(x-1)$ and its graph $C_a$ depending on the parameter $a$. We need to:

a) Find the value(s) of $a$ such that the graph $C_a$ is tangent to the x-axis.

b) Study the variation and sketch the graph of the function for the value(s) of $a$ found in part (a).

2. Solution Steps

a) Find the value(s) of $a$ such that the graph $C_a$ is tangent to the x-axis.

If the graph $C_a$ is tangent to the x-axis, there exists a value $x_0$ such that $f_a(x_0) = 0$ and $f'_a(x_0) = 0$.

First, let's find the derivative of $f_a(x)$:

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

$f'_a(x) = 3x^2 - a$

Now, we have the following system of equations:

1. $f_a(x_0) = x_0^3 - 1 - a(x_0 - 1) = 0$

2. $f'_a(x_0) = 3x_0^2 - a = 0$

From equation (2), we can express $a$ in terms of $x_0$:

$a = 3x_0^2$

Substituting this expression for $a$ into equation (1):

$x_0^3 - 1 - (3x_0^2)(x_0 - 1) = 0$

$x_0^3 - 1 - 3x_0^3 + 3x_0^2 = 0$

$-2x_0^3 + 3x_0^2 - 1 = 0$

$2x_0^3 - 3x_0^2 + 1 = 0$

We can observe that $x_0 = 1$ is a solution:

$2(1)^3 - 3(1)^2 + 1 = 2 - 3 + 1 = 0$

So, $(x_0 - 1)$ is a factor. Now we can perform polynomial division:

$(2x_0^3 - 3x_0^2 + 1) / (x_0 - 1) = 2x_0^2 - x_0 - 1$

So, $2x_0^3 - 3x_0^2 + 1 = (x_0 - 1)(2x_0^2 - x_0 - 1) = 0$

Now, we solve the quadratic equation $2x_0^2 - x_0 - 1 = 0$:

We can factor this quadratic as $(2x_0 + 1)(x_0 - 1) = 0$.

The solutions are $x_0 = 1$ and $x_0 = -\frac{1}{2}$.

Thus, the roots are $x_0 = 1$ (double root) and $x_0 = -\frac{1}{2}$.

Now, we find the corresponding values of $a$:

For $x_0 = 1$, $a = 3(1)^2 = 3$.

For $x_0 = -\frac{1}{2}$, $a = 3(-\frac{1}{2})^2 = 3(\frac{1}{4}) = \frac{3}{4}$.

When $a = 3$, $f_3(x) = x^3 - 1 - 3(x-1) = x^3 - 3x + 2 = (x-1)^2(x+2)$. Since $(x-1)^2$ is a factor, the graph touches the x-axis at $x=1$.

When $a = \frac{3}{4}$, $f_{3/4}(x) = x^3 - 1 - \frac{3}{4}(x-1) = x^3 - \frac{3}{4}x - \frac{1}{4}$. Since $x_0 = -\frac{1}{2}$, we know that $f_{3/4}(-\frac{1}{2}) = 0$.

b) Study the variation and sketch the graph of the function for the value(s) of $a$ found in part (a).

Case 1: $a = 3$

$f_3(x) = x^3 - 3x + 2 = (x-1)^2(x+2)$

$f'_3(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$

Critical points: $x = -1$ and $x = 1$.

$f_3(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$

$f_3(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$

As $x \rightarrow \infty$, $f_3(x) \rightarrow \infty$

As $x \rightarrow -\infty$, $f_3(x) \rightarrow -\infty$

Case 2: $a = \frac{3}{4}$

$f_{3/4}(x) = x^3 - \frac{3}{4}x - \frac{1}{4}$

$f'_{3/4}(x) = 3x^2 - \frac{3}{4} = \frac{3}{4}(4x^2 - 1) = \frac{3}{4}(2x - 1)(2x + 1)$

Critical points: $x = -\frac{1}{2}$ and $x = \frac{1}{2}$

$f_{3/4}(-\frac{1}{2}) = (-\frac{1}{2})^3 - \frac{3}{4}(-\frac{1}{2}) - \frac{1}{4} = -\frac{1}{8} + \frac{3}{8} - \frac{2}{8} = 0$

$f_{3/4}(\frac{1}{2}) = (\frac{1}{2})^3 - \frac{3}{4}(\frac{1}{2}) - \frac{1}{4} = \frac{1}{8} - \frac{3}{8} - \frac{2}{8} = -\frac{4}{8} = -\frac{1}{2}$

As $x \rightarrow \infty$, $f_{3/4}(x) \rightarrow \infty$

As $x \rightarrow -\infty$, $f_{3/4}(x) \rightarrow -\infty$

3. Final Answer

a) $a = 3$ and $a = \frac{3}{4}$

b) Case 1: $a=3$, $f_3(x) = x^3 - 3x + 2$. Tangent at $x=1$, local max at $x=-1$.

Case 2: $a = \frac{3}{4}$, $f_{3/4}(x) = x^3 - \frac{3}{4}x - \frac{1}{4}$. Tangent at $x=-\frac{1}{2}$, local min at $x=\frac{1}{2}$.