Math Problem Statement

Soit f la fonction définie sur IR+ par : f(x) = x - 2√x + 1

Solution

Let's go step by step in solving the mathematical tasks outlined in the image.

Given Function:

The function is defined as f(x)=x2x+1f(x) = x - 2\sqrt{x} + 1, with the domain xR+x \in \mathbb{R}^+.

1) Verifier que f(x)=(x1)2f(x) = (\sqrt{x} - 1)^2

We need to verify that the given expression for f(x)f(x) can be rewritten as (x1)2(\sqrt{x} - 1)^2.

  • Starting with f(x)=x2x+1f(x) = x - 2\sqrt{x} + 1: (x1)2=(x)22(x)(1)+(1)2=x2x+1.(\sqrt{x} - 1)^2 = (\sqrt{x})^2 - 2(\sqrt{x})(1) + (1)^2 = x - 2\sqrt{x} + 1.
  • Thus, f(x)=(x1)2f(x) = (\sqrt{x} - 1)^2, which is verified.

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

To calculate this limit: f(x)=x2x+1.f(x) = x - 2\sqrt{x} + 1.

As x+x \to +\infty:

  • x+x \to +\infty,
  • x+\sqrt{x} \to +\infty, so 2x-2\sqrt{x} \to -\infty.

Thus, the behavior of f(x)f(x) at infinity is dominated by xx, so: limx+f(x)=+.\lim_{x \to +\infty} f(x) = +\infty.

3) Continuité de la fonction ff sur R+\mathbb{R}^+

Since the function is composed of basic continuous functions (polynomials and square roots), and there is no discontinuity in f(x)=x2x+1f(x) = x - 2\sqrt{x} + 1 for x0x \geq 0, the function is continuous on R+\mathbb{R}^+.

4) Dérivabilité de la fonction à droite de 0 et interprétation géométrique

To check the differentiability, we first calculate the derivative of f(x)f(x) for x>0x > 0.

f(x)=ddx(x2x+1)=122x=11x.f'(x) = \frac{d}{dx}(x - 2\sqrt{x} + 1) = 1 - \frac{2}{2\sqrt{x}} = 1 - \frac{1}{\sqrt{x}}.

Next, evaluate the right-hand derivative at x=0x = 0 using the limit: f+(0)=limx0+f(x)f(0)x0=limx0+(x2x+1)1x=limx0+x2xx.f'_+(0) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{(x - 2\sqrt{x} + 1) - 1}{x} = \lim_{x \to 0^+} \frac{x - 2\sqrt{x}}{x}.

Simplifying: f+(0)=limx0+(12x).f'_+(0) = \lim_{x \to 0^+} \left(1 - \frac{2}{\sqrt{x}}\right).

As x0+x \to 0^+, 2x\frac{2}{\sqrt{x}} \to \infty, so: f+(0)=.f'_+(0) = -\infty. Thus, the function is not differentiable at 0.


The remaining questions involve further exploration of the function, its variations, tangents, and inverse. Would you like me to continue solving them or explain any part in more detail?

Here are 5 related questions for deeper understanding:

  1. How do we analyze the global behavior of f(x)f(x) from a graph perspective?
  2. Can the non-differentiability at 0 be interpreted geometrically?
  3. What methods can we use to find the equation of a tangent line at any given point?
  4. How can we determine if ff is invertible on a specific interval?
  5. What are the conditions under which a function's inverse is differentiable?

Tip: For functions with square roots, always check for potential issues in differentiability and domain boundaries.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Differentiability
Inverse Functions

Formulas

f(x) = (√x - 1)^2
lim_{x -> +∞} f(x) = +∞
f'(x) = 1 - 1/√x
f'_+(0) = lim_{x -> 0^+} (x - 2√x)/x

Theorems

Limit Definition
Continuity Theorem
Derivative Definition

Suitable Grade Level

Undergraduate Calculus