The problem is based on the function $f(x) = 1 - x - e^x$. It involves calculating limits, finding the derivative, analyzing the relative position of the curve with respect to its oblique asymptote, showing the existence of an inverse function, finding the value of the inverse function at 0, and finding the intersection points with the coordinate axes. Finally, it requires plotting the function, its inverse and calculating the area.

1. Problem Description

The problem is based on the function $f(x) = 1 - x - e^x$. It involves calculating limits, finding the derivative, analyzing the relative position of the curve with respect to its oblique asymptote, showing the existence of an inverse function, finding the value of the inverse function at 0, and finding the intersection points with the coordinate axes. Finally, it requires plotting the function, its inverse and calculating the area.

2. Solution Steps

1a) Calculate the limits:

$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (1 - x - e^x)$. As $x \to -\infty$, $e^x \to 0$ and $-x \to \infty$, so $\lim_{x \to -\infty} f(x) = \infty$.

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (1 - x - e^x)$. As $x \to +\infty$, $-x \to -\infty$ and $-e^x \to -\infty$, so $\lim_{x \to +\infty} f(x) = -\infty$.

$\lim_{x \to +\infty} (f(x) - (1-x)) = \lim_{x \to +\infty} (1 - x - e^x - (1-x)) = \lim_{x \to +\infty} (-e^x) = -\infty$.

$\lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{1 - x - e^x}{x}$. We can apply L'Hopital's rule since it is in the indeterminate form $\frac{0}{0}$. Differentiating the numerator and denominator, we get $\lim_{x \to 0} \frac{-1 - e^x}{1} = -1 - e^0 = -1 - 1 = -2$.

1b) Calculate $f'(x)$ and determine its sign to create the table of variations (TV).

$f'(x) = -1 - e^x$.

Since $e^x > 0$ for all $x \in \mathbb{R}$, $-1 - e^x < -1 < 0$. Therefore, $f'(x) < 0$ for all $x \in \mathbb{R}$. The function $f(x)$ is strictly decreasing.

Table of Variations:

x | -inf | +inf

-------|--------|-------

f'(x) | - | -

-------|--------|-------

f(x) | +inf | -inf

1c) Study the relative position of $C_f$ with respect to its oblique asymptote $\Delta$ of equation $y=1-x$.

The expression $f(x) - (1-x) = 1-x-e^x - (1-x) = -e^x$.

Since $e^x > 0$ for all $x$, $-e^x < 0$. Therefore, $f(x) - (1-x) < 0$, which means $f(x) < 1-x$. The curve $C_f$ is below the line $\Delta$ for all $x$. $\Delta$ is the asymptote for the function.

2a) Show that $f$ has a reciprocal function $f^{-1}$.

Since $f'(x) = -1-e^x < 0$, $f(x)$ is strictly decreasing and continuous over $\mathbb{R}$. Therefore, $f(x)$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$, so $f$ has an inverse function $f^{-1}$.

2b) Calculate $f(0)$ and deduce the sign of $f$.

$f(0) = 1 - 0 - e^0 = 1 - 1 = 0$.

Since $f$ is strictly decreasing and $f(0) = 0$, we have:

- if $x < 0$, $f(x) > 0$

- if $x > 0$, $f(x) < 0$

3a) Show that $f^{-1}(0) = -\frac{1}{2}$.

This statement is incorrect. From 2b) we have $f(0)=0$. This means that $f^{-1}(0) = 0$, not $-\frac{1}{2}$.

The equation of the tangent line to $C_f$ at the point with abscissa $x_0$ is given by:

$y = f'(x_0) (x - x_0) + f(x_0)$

Since $f(0) = 0$, we can determine the equation of the tangent line $T_1$ at $x_0 = 0$:

$f'(0) = -1 - e^0 = -1 - 1 = -2$

$T_1: y = -2(x - 0) + 0 = -2x$

The problem asks for the equation of the tangent lines $T_1$ and $T_2$ to $C_f$. However, the problem does not mention anything about the points of abscissa $T_1$ and $T_2$, so we can only calculate $T_1$, where $x_0 = 0$.

3b) Specify the point(s) where $C_f$ intersects the coordinate axes.

$C_f$ intersects the y-axis when $x = 0$. We already know $f(0) = 0$. So the curve intersects the y-axis at $(0, 0)$.

$C_f$ intersects the x-axis when $f(x) = 0$. We know that $f(0) = 0$. So the curve intersects the x-axis at $(0, 0)$.

3c) Trace $C_f$ and $C_{f^{-1}}$ on the same orthonormal coordinate system (O; I,J) with unit 2cm. Shade and calculate in cm² the area A of $C_f$, the asymptote $\Delta$ and the lines of equation.

Since the image does not provide specific lines, I cannot compute the area.

3. Final Answer

1a) $\lim_{x \to -\infty} f(x) = \infty$, $\lim_{x \to +\infty} f(x) = -\infty$, $\lim_{x \to +\infty} (f(x) - (1-x)) = -\infty$, $\lim_{x \to 0} \frac{f(x)}{x} = -2$

1b) $f'(x) = -1 - e^x$, f(x) is strictly decreasing.

1c) $C_f$ is below the line $\Delta$ for all $x$.

2a) $f$ has an inverse function $f^{-1}$.

2b) $f(0) = 0$, if $x < 0$, $f(x) > 0$, if $x > 0$, $f(x) < 0$

3a) $f^{-1}(0) = 0$, $T_1: y = -2x$

3b) $(0, 0)$

3c) Cannot answer the question regarding calculating the area.