Step 1: Solve the geometry problem. 1) Write the relation that links $x$ to the number $f(x)$. The point $M$ lies on a semicircle with diameter $[AB]$. According to Thales' theorem, the angle $\angle AMB$ is a right angle. Therefore, $\triangle AMB$ is a right-angled triangle at $M$. Given $AM = x$ and $BM = f(x)$, and the diameter $AB = 2$. Using the Pythagorean theorem: $$AM^2 + BM^2 = AB^2$$ $$x^2 + (f(x))^2 = 2^2$$ $$x^2 + (f(x))^2 = 4$$ To express $f(x)$ in terms of $x$: $$(f(x))^2 = 4 - x^2$$ Since $BM$ is a distance, $f(x)$ must be non-negative. $$f(x) = \sqrt{4 - x^2}$$ The relation is $\boxed{f(x) = \sqrt{4 - x^2}}$. 2) a- Calculate the distance $BM$ in the case $AM = \frac{2}{3}$. Given $AM = x = \frac{2}{3}$. Substitute this value into the function $f(x)$: $$BM = f\left(\frac{2}{3}\right) = \sqrt{4 - \left(\frac{2}{3}\right)^2}$$ $$BM = \sqrt{4 - \frac{4}{9}}$$ $$BM = \sqrt{\frac{36}{9} - \frac{4}{9}}$$ $$BM = \sqrt{\frac{32}{9}}$$ $$BM = \frac{\sqrt{32}}{\sqrt{9}}$$ $$BM = \frac{\sqrt{16 \times 2}}{3}$$ $$BM = \frac{4\sqrt{2}}{3}$$ The distance $BM$ is $\boxed{\frac{4\sqrt{2}}{3}}$. 2) b- What are the numbers $x$ that have an image by the function $f$? This set of numbers is called the domain of definition of the function $f$ and is denoted by $D_f$. The function is $f(x) = \sqrt{4 - x^2}$. For $f(x)$ to be defined, the expression under the square root must be non-negative: $$4 - x^2 \ge 0$$ $$x^2 \le 4$$ This implies $-2 \le x \le 2$. Additionally, $x$ represents the distance $AM$, so $x$ must be non-negative ($x \ge 0$). Combining $x \ge 0$ and $-2 \le x \le 2$, the possible values for $x$ are $0 \le x \le 2$. The domain of definition $D_f$ is $\boxed{[0, 2]}$. Step 2: Determine the domain of definition for the given functions. 3) Determine the domain of definition for the following functions: • For $f_1: x \mapsto \frac{1}{x}$: The denominator cannot be zero. $$x \ne 0$$ The domain is $D_{f_1} = \boxed{\mathbb{R} \setminus \{0\}}$. • For $f_2: x \mapsto \sqrt{x}$: The expression under the square root must be non-negative. $$x \ge 0$$ The domain is $D_{f_2} = \boxed{[0, +\infty)}$. • For $f_3: x \mapsto \sqrt{x+2}$: The expression under the square root must be non-negative. $$x+2 \ge 0$$ $$x \ge -2$$ The domain is $D_{f_3} = \boxed{[-2, +\infty)}$. • For $f_4: x \mapsto 3x^2+x+1$: This is a polynomial function, which is defined for all real numbers. The domain is $D_{f_4} = \boxed{\mathbb{R}}$. • For $f_5: x \mapsto x\sqrt{x^2-1}$: The expression under the square root must be non-negative. $$x^2-1 \ge 0$$ $$x^2 \ge 1$$ This implies $x \le -1$ or $x \ge 1$. The domain is $D_{f_5} = \boxed{(-\infty, -1] \cup [1, +\infty)}$. • For $f_6: x \mapsto \frac{\sqrt{2x+1}}{x}$: Two conditions must be met: 1. The expression under the square root must be non-negative: $2x+1 \ge 0 \implies 2x \ge -1 \implies x \ge -\frac{1}{2}$. 2. The denominator cannot be zero: $x \ne 0$. Combining these conditions, $x$ must be greater than or equal to $-\frac{1}{2}$ and not equal to $0$. The domain is $D_{f_6} = \boxed{\left[-\frac{1}{2}, 0\right) \cup (0, +\infty)}$. • For $f_7: x \mapsto \frac{\sqrt{x^2+x-2}}{x-1}$: Two conditions must be met: 1. The expression under the square root must be non-negative: $x^2+x-2 \ge 0$. Factor the quadratic: $(x+2)(x-1) \ge 0$. The roots are $x=-2$ and $x=1$. The parabola opens upwards, so the expression is non-negative when $x \le -2$ or $x \ge 1$. 2. The denominator cannot be zero: $x-1 \ne 0 \implies x \ne 1$. Combining these conditions: $x \le -2$ or $x \ge 1$, and $x \ne 1$. This means $x \le -2$ or $x > 1$. The domain is $D_{f_7} = \boxed{(-\infty, -2] \cup (1, +\infty)}$.