https://file.aiquickdraw.com/mathbot/file/17295306990858diklf9v.jpg

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!

Soit f la fonction définie sur IR^+ par : f(x) = x - 2√x + 1.
1) Vérifier que f(x) = (√x - 1)^2
2) Calculer lim (x→+∞) f(x)
3) Etudier la continuité de la fonction f sur IR^+
4) a) Etudier la dérivabilité de la fonction à droite de 0, puis interpréter géométriquement le résultat obtenu.

Explore the properties of the function f(x) = x - 2√x + 1, including its limits as x approaches infinity, continuity on IR^+, and differentiability at x = 0. A detailed breakdown of key algebraic and calculus concepts.

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