The image contains a set of mathematical problems related to a function $f(x)$. The tasks include calculating the derivative $f'(x)$ and determining its sign, creating a table of variations (TV) for $f$, studying the relative position of the curve $C_f$ with respect to its oblique asymptote $(\Delta)$, proving that $f$ has an inverse function $f^{-1}$ and giving its table of variations, calculating $f(0)$ and deducing the sign of $f$, showing that $f^{-1}(0) = -\frac{1}{2}$ and finding the equations of the tangents $T_1$ and $T_2$ to $C_f$ at the points with abscissa 0 and 1, finding the points where $C_f$ intersects the axes, plotting $C_f$ and $C_{f^{-1}}$ on the same orthonormal coordinate system with unit length 2cm, and calculating the area enclosed by $C_f$, $(\Delta)$, and the lines $x = -1$ and $x$.

1. Problem Description

The image contains a set of mathematical problems related to a function $f(x)$. The tasks include calculating the derivative $f'(x)$ and determining its sign, creating a table of variations (TV) for $f$, studying the relative position of the curve $C_f$ with respect to its oblique asymptote $(\Delta)$, proving that $f$ has an inverse function $f^{-1}$ and giving its table of variations, calculating $f(0)$ and deducing the sign of $f$, showing that $f^{-1}(0) = -\frac{1}{2}$ and finding the equations of the tangents $T_1$ and $T_2$ to $C_f$ at the points with abscissa 0 and 1, finding the points where $C_f$ intersects the axes, plotting $C_f$ and $C_{f^{-1}}$ on the same orthonormal coordinate system with unit length 2cm, and calculating the area enclosed by $C_f$, $(\Delta)$, and the lines $x = -1$ and $x$.

2. Solution Steps

Since the function $f(x)$ is not given explicitly, I cannot solve the problems numerically. However, I will outline the general steps required to solve each part:

a) To calculate $f'(x)$, apply the rules of differentiation depending on the specific form of $f(x)$ (e.g., power rule, product rule, chain rule). To determine the sign of $f'(x)$, find the critical points where $f'(x) = 0$ or is undefined. Create a sign chart using test values in the intervals defined by these critical points. The table of variations (TV) for $f$ can then be constructed using the sign of $f'(x)$ to indicate where $f$ is increasing or decreasing.

b) To study the relative position of $C_f$ with respect to its oblique asymptote $(\Delta)$, find the equation of the oblique asymptote $(\Delta)$. Then, study the sign of the difference $f(x) - (\text{equation of } \Delta)$. If $f(x) - (\text{equation of } \Delta) > 0$, $C_f$ is above $(\Delta)$. If $f(x) - (\text{equation of } \Delta) < 0$, $C_f$ is below $(\Delta)$. If $f(x) - (\text{equation of } \Delta) = 0$, $C_f$ intersects $(\Delta)$.

2a) To prove that $f$ has an inverse function $f^{-1}$, show that $f$ is strictly monotonic (either strictly increasing or strictly decreasing) over its domain. Then, you can determine the TV of the inverse function as the range of $f$ over the corresponding intervals.

b) To calculate $f(0)$, substitute $x=0$ into the expression for $f(x)$. The sign of $f(0)$ is simply whether the value is positive or negative.

3a) To show that $f^{-1}(0) = -\frac{1}{2}$, you would need to know what $f(x)$ is. This implies that $f(-\frac{1}{2}) = 0$. To find the equation of the tangent $T_1$ to $C_f$ at $x=0$, use the formula $y = f'(0)(x-0) + f(0)$. To find the equation of the tangent $T_2$ to $C_f$ at $x=1$, use the formula $y = f'(1)(x-1) + f(1)$.

b) To find the points where $C_f$ intersects the axes, set $x=0$ to find the $y$-intercept (which is already calculated as $f(0)$), and set $f(x) = 0$ to find the $x$-intercept(s).

c) Plot $C_f$ and $C_{f^{-1}}$ on the same coordinate system. Remember that $C_{f^{-1}}$ is the reflection of $C_f$ over the line $y=x$.

pa4a) To calculate the area enclosed by $C_f$, $(\Delta)$, and the lines $x = -1$ and $x$, integrate the absolute value of the difference between $f(x)$ and the equation of the asymptote $(\Delta)$ from $x=-1$ to $x$:

$Area = \int_{-1}^{x} |f(t) - (\text{equation of } \Delta)| dt$.

Since the scale is 2cm per unit, the area calculated by the integral will be in units squared. You need to multiply this result by $(2 \text{ cm})^2 = 4 \text{ cm}^2$ to get the area in cm$^2$.

3. Final Answer

Due to the lack of the explicit form of $f(x)$, I cannot provide numerical solutions. The above are the steps needed to derive the solution of the problem.