https://file.aiquickdraw.com/mathbot/file/1726075076886p1qknat1.jpg

Let's go through the exercises step by step:

### Exercice 1:

The graph shows the function \( f \), defined and differentiable over the set of real numbers, with the following conditions:
- Point \( G(-2; 5) \) and point \( H(0; 1) \) are on the curve.
- The tangents at points \( G \) and \( H \) are horizontal.

#### Questions:

1. **Determine \( f(0) \) and \( f(-2) \), justifying your answers.**

   - From the graph, at point \( H(0; 1) \), we have \( f(0) = 1 \).
   - At point \( G(-2; 5) \), we have \( f(-2) = 5 \).

2. **Determine \( f'(0) \) and \( f'(-2) \), justifying your answers.**

   - The tangents at points \( G \) and \( H \) are horizontal, which means the derivative at these points is zero:
     - \( f'(-2) = 0 \)
     - \( f'(0) = 0 \)

3. **The function \( f \) is defined by \( f(x) = ax^3 + bx^2 + cx + d \). Determine \( a \), \( b \), \( c \), and \( d \).**

   - **(a)** Find \( f'(x) \) in terms of \( a, b, c, d \):
     \[
     f'(x) = 3ax^2 + 2bx + c
     \]
   
   - **(b)** Determine \( c \) and \( d \):
     - Using the point \( H(0; 1) \): \( f(0) = d = 1 \), so \( d = 1 \).
     - Using the point \( G(-2; 5) \): Substitute into \( f(-2) = 8a - 4b - 2c + d = 5 \). Given \( d = 1 \), we get:
       \[
       8a - 4b - 2c + 1 = 5 \implies 8a - 4b - 2c = 4 \implies 4a - 2b - c = 2
       \]

   - **(c)** Find two equations for \( a \) and \( b \):
     - Using \( f'(0) = 0 \): From \( f'(x) = 3ax^2 + 2bx + c \), substitute \( x = 0 \):
       \[
       c = 0 \implies c = 0
       \]
     - From \( f'(-2) = 0 \):
       \[
       f'(-2) = 12a - 4b + c = 0
       \]
     - Since \( c = 0 \), the equation becomes:
       \[
       12a - 4b = 0 \implies 3a = b
       \]

   - **(d)** Determine \( a \) and \( b \), and thus \( f(x) \):
     - Using \( 4a - 2b = 2 \) and \( b = 3a \):
       \[
       4a - 2(3a) = 2 \implies 4a - 6a = 2 \implies -2a = 2 \implies a = -1
       \]
       \[
       b = 3a = -3
       \]
     - Therefore, \( f(x) = -x^3 - 3x^2 + 1 \).

### Exercice 2:

Let \( f(x) = (x^2 + 2x + 1)e^{-x} \).

#### Questions:

1. **Show in two different ways that \( f(x) \geq 0 \) for all real \( x \).**

   - Rewrite \( f(x) = (x + 1)^2 e^{-x} \). Since \( (x+1)^2 \geq 0 \) and \( e^{-x} > 0 \), the product is non-negative.

2. **Show that \( f \) is differentiable over \( \mathbb{R} \).**
   
   - The product and chain rule indicate \( f \) is differentiable since both components are differentiable.

3. **Study the sign of \( f' \) and deduce the variations of \( f \) over \( \mathbb{R} \).**
   
   - Given \( f'(x) = (-x^2 + 1)e^{-x} \), solve \( f'(x) = 0 \) and analyze the intervals to determine increasing or decreasing behavior.

4. **Construct the graph of \( f(x) \).**

Would you like more details on any specific part, or help with graph construction? Here are some further questions to explore:

1. How does differentiability relate to the continuity of a function?
2. How can you confirm the results of the derivative graphically?
3. What are the implications of horizontal tangents for local maxima or minima?
4. How does the behavior of the exponential function affect the function \( f(x) \)?
5. How can the second derivative test be used to confirm concavity?

**Tip:** To sketch functions accurately, always find critical points, inflection points, and asymptotic behavior!

The curve below represents a function f defined and differentiable over all real numbers. The points G(-2; 5) and H(0; 1) are on the graph of f, and the tangents at these points are horizontal.

Solve a cubic function problem by finding derivatives and analyzing points where the tangent is horizontal. This problem involves polynomial functions, derivatives, and differentiation concepts suitable for high school calculus students.

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