Let's break down the tasks from the image step by step:

1) Vérifier que $f(x) = ( \sqrt{x} - 1 )^2$

The function given is $f(x) = x - 2\sqrt{x} + 1$.

We are asked to verify that $f(x)$ can also be written as $( \sqrt{x} - 1 )^2$.

Verification:

First, expand $( \sqrt{x} - 1 )^2$:

$( \sqrt{x} - 1 )^2 = ( \sqrt{x} - 1 )( \sqrt{x} - 1 ) = x - 2\sqrt{x} + 1$

Since this is the same as the given expression for $f(x)$, the verification is correct.

2) Calculer $\lim_{x \to +\infty} f(x)$

We need to calculate the limit as $x$ approaches infinity for $f(x) = ( \sqrt{x} - 1 )^2$.

$f(x) = ( \sqrt{x} - 1 )^2 = x - 2\sqrt{x} + 1$

When $x \to +\infty$:

• $x \to \infty$
• $2\sqrt{x} \to \infty$

Thus, the leading term $x$ dominates, and we get:

$\lim_{x \to +\infty} f(x) = \infty$

3) Étudier la continuité de la fonction $f$ sur $\mathbb{R}^+$

The function $f(x) = x - 2\sqrt{x} + 1$ is a composition of continuous functions:

• $x$ is continuous.
• $\sqrt{x}$ is continuous on $\mathbb{R}^+$.
• The operations of addition, subtraction, and squaring preserve continuity.

Thus, $f(x)$ is continuous on $\mathbb{R}^+$.

4a) Étudier la dérivabilité de la fonction à droite de 0

We now need to study the differentiability of $f(x)$ at 0 from the right.

Step 1: Compute the derivative for $x > 0$

We first find the derivative of $f(x) = ( \sqrt{x} - 1 )^2$:

Using the chain rule: $f'(x) = 2( \sqrt{x} - 1 ) \cdot \frac{1}{2\sqrt{x}} = \frac{\sqrt{x} - 1}{\sqrt{x}}$

For $x = 0^+$: $f'(0^+) = \lim_{x \to 0^+} \frac{\sqrt{x} - 1}{\sqrt{x}} = \lim_{x \to 0^+} \left( 1 - \frac{1}{\sqrt{x}} \right) = -\infty$

Thus, $f'(0^+)$ tends to $-\infty$, meaning the function has a vertical tangent at $x = 0$.

Step 2: Interpretation (Géométriquement le résultat obtenu)

The fact that $f'(0^+) = -\infty$ indicates that the function has a very steep downward slope as it approaches $x = 0$ from the right. This suggests a vertical tangent line at $x = 0$.

Summary

• 1) The expression $f(x) = ( \sqrt{x} - 1 )^2$ is verified.
• 2) $\lim_{x \to +\infty} f(x) = \infty$.
• 3) The function $f$ is continuous on $\mathbb{R}^+$.
• 4a) The derivative $f'(0^+) = -\infty$, indicating a vertical tangent at $x = 0$.

Do you want more details on any of the steps? Here are some related questions for further exploration:

1. What are the conditions for differentiability at a point?
2. How can we interpret limits that approach infinity in real-world contexts?
3. Can we generalize the behavior of $f(x)$ for very large values of $x$?
4. What are some other examples of functions with vertical tangents?
5. How can we rigorously prove the continuity of a function using epsilon-delta definitions?

Tip: Always check for continuity before testing differentiability, as differentiability implies continuity but not vice versa!