We are given a function $f(x)$ defined piecewise as: $f(x) = x + \sqrt{1-x^2}$ for $x \in [-1, 1]$ $f(x) = \frac{1+\sqrt{x^2-1}}{x}$ for $x \in [-1, 0[ \cup ]0, 1]$ Part A asks to: 1. Determine the domain of $f$.

1. Problem Description

We are given a function $f(x)$ defined piecewise as:

$f(x) = x + \sqrt{1-x^2}$ for $x \in [-1, 1]$

$f(x) = \frac{1+\sqrt{x^2-1}}{x}$ for $x \in [-1, 0[ \cup ]0, 1]$

Part A asks to:

1. Determine the domain of $f$.

2. Study the continuity of $f$ at $x = -1$ and $x = 1$.

3. Study the differentiability of $f$ at $x = -1$ and $x = 1$. Deduce the geometric interpretation.

4. Study the variations of $f$ and construct the variation table.

5. Study the infinite branch of the curve (C) of $f$.

6. Draw the curve (C) of $f$.

Part B defines a new function $h(x) = -f(x)$, the restriction of $f$ on the interval $I = [-1, 1]$. It asks to:

1. Construct the variation table of $h$.

2. Draw the curve (C') of $h$.

2. Solution Steps

Part A:

1. Domain of $f$:

The domain is given in the definition: $D_f = [-1, 0[ \cup ]0, 1] \cup [-1, 1] = [-1, 0[ \cup ]0, 1] \cup \{-1\} \cup \{1\} \cup ]-1,1[ = [-1, 1]$ excluding $0$. Since the second function's interval is $[-1, 0[ \cup ]0, 1]$, and the first function's is $[-1,1]$, the entire domain is the union of both, which gives $[-1, 0[ \cup ]0, 1] \cup \{-1\} \cup \{1\} \cup ]-1,1[= [-1, 1] \setminus \{0\}$. However, the first function is defined as $x+\sqrt{1-x^2}$ for $x\in[-1,1]$, so it is defined at $x=0$ and equals $0+\sqrt{1-0^2} = 1$. We are given that for $x$ in $]-1, 0[\cup]0, 1[$, $f(x) = \frac{1 + \sqrt{x^2-1}}{x}$. For this second formula to make sense, $x^2-1 \ge 0$, i.e. $|x|\ge 1$. Thus we must have $x \in (-\infty, -1] \cup [1, \infty)$. Given the restriction that $x$ is in $]-1,0[\cup]0,1[$, there is no overlapping region. Therefore, the second part doesn't exist.

So $f(x) = x + \sqrt{1-x^2}$ for $x\in[-1,1]$

Then $D_f = [-1, 1]$.

2. Continuity of $f$ at $x = -1$ and $x = 1$:

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

At $x = -1$: $f(-1) = -1 + \sqrt{1 - (-1)^2} = -1$. $\lim_{x \to -1^+} f(x) = -1$, thus $f$ is continuous at $x=-1$.

At $x = 1$: $f(1) = 1 + \sqrt{1 - 1^2} = 1$. $\lim_{x \to 1^-} f(x) = 1$, thus $f$ is continuous at $x=1$.

3. Differentiability of $f$ at $x = -1$ and $x = 1$:

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

At $x = -1$: $\lim_{x \to -1^+} f'(x) = \lim_{x \to -1^+} 1 - \frac{x}{\sqrt{1 - x^2}} = +\infty$. $f$ is not differentiable at $x = -1$.

At $x = 1$: $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 1 - \frac{x}{\sqrt{1 - x^2}} = -\infty$. $f$ is not differentiable at $x = 1$.

Geometric interpretation: At $x=-1$ and $x=1$, the curve has vertical tangents.

4. Variations of $f$:

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

$f'(x) = 0 \implies 1 = \frac{x}{\sqrt{1 - x^2}} \implies \sqrt{1 - x^2} = x$.

$1 - x^2 = x^2 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$.

Since $\sqrt{1 - x^2} = x$, $x$ must be positive, thus $x = \frac{1}{\sqrt{2}}$.

If $x < \frac{1}{\sqrt{2}}$, $f'(x) > 0$. If $x > \frac{1}{\sqrt{2}}$, $f'(x) < 0$.

$f(\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} + \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}} + \sqrt{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Variation table:

x | -1 1/sqrt(2) 1

------------------------

f'(x) | + 0 -

------------------------

f(x) | -1 increasing sqrt(2) decreasing 1

5. Infinite branch of the curve:

The domain is $[-1, 1]$. Thus, there is no infinite branch.

6. Draw the curve (C) of $f$: The graph is a semi-circle translated to the left.

Part B:

$h(x) = -f(x) = -x - \sqrt{1 - x^2}$ for $x \in [-1, 1]$.

1. Variation table of $h$:

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

$h'(x) = 0 \implies -1 = - \frac{x}{\sqrt{1 - x^2}} \implies \sqrt{1 - x^2} = -x$.

$1 - x^2 = x^2 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$.

Since $\sqrt{1 - x^2} = -x$, $x$ must be negative, thus $x = -\frac{1}{\sqrt{2}}$.

If $x < -\frac{1}{\sqrt{2}}$, $h'(x) < 0$. If $x > -\frac{1}{\sqrt{2}}$, $h'(x) > 0$.

$h(-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} - \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}} - \sqrt{\frac{1}{2}} = 0$.

Variation table:

x | -1 -1/sqrt(2) 1

------------------------

h'(x) | - 0 +

------------------------

h(x) | 1 decreasing 0 increasing -1

2. Draw the curve (C') of $h$: The curve is the reflection across the x-axis of the curve of $f$.

3. Final Answer

Part A:

1. $D_f = [-1, 1]$

2. $f$ is continuous at $x=-1$ and $x=1$.

3. $f$ is not differentiable at $x=-1$ and $x=1$. The curve has vertical tangents at $x=-1$ and $x=1$.

4. Variation table of $f$:

x | -1 1/sqrt(2) 1

------------------------

f'(x) | + 0 -

------------------------

f(x) | -1 increasing sqrt(2) decreasing 1

5. No infinite branch.

6. Graph of $f(x)$ is a semi-circle translated to the left.

Part B:

1. Variation table of $h$:

x | -1 -1/sqrt(2) 1

------------------------

h'(x) | - 0 +

------------------------

h(x) | 1 decreasing 0 increasing -1

2. Graph of $h(x)$ is the reflection across the x-axis of the curve of $f$.