This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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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 AMB is a right angle. Therefore, 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 f(x) = sqrt(4 - x^2). 2) a- Calculate the distance BM in the case AM = (2)/(3). Given AM = x = (2)/(3). Substitute this value into the function f(x): BM = f((2)/(3)) = sqrt(4 - ((2)/(3))^2) BM = sqrt(4 - (4)/(9)) BM = sqrt((36)/(9) - (4)/(9)) BM = sqrt((32)/(9)) BM = sqrt(32)sqrt(9) BM = sqrt(16 × 2)3 BM = 4sqrt(2)3 The distance BM is 4sqrt(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 0 x^2 4 This implies -2 x 2. Additionally, x represents the distance AM, so x must be non-negative (x 0). Combining x 0 and -2 x 2, the possible values for x are 0 x 2. The domain of definition D_f is [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 (1)/(x): The denominator cannot be zero. x 0 The domain is D_f_1 = R \0\. • For f_2: x sqrt(x): The expression under the square root must be non-negative. x 0 The domain is D_f_2 = [0, +). • For f_3: x sqrt(x+2): The expression under the square root must be non-negative. x+2 0 x -2 The domain is D_f_3 = [-2, +). • For f_4: x 3x^2+x+1: This is a polynomial function, which is defined for all real numbers. The domain is D_f_4 = R. • For f_5: x xsqrt(x^2-1): The expression under the square root must be non-negative. x^2-1 0 x^2 1 This implies x -1 or x 1. The domain is D_f_5 = (-, -1] [1, +). • For f_6: x sqrt(2x+1)x: Two conditions must be met: 1. The expression under the square root must be non-negative: 2x+1 0 2x -1 x -(1)/(2). 2. The denominator cannot be zero: x 0. Combining these conditions, x must be greater than or equal to -(1)/(2) and not equal to 0. The domain is D_f_6 = [-(1)/(2), 0) (0, +). • For f_7: x 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 0. Factor the quadratic: (x+2)(x-1) 0. The roots are x=-2 and x=1. The parabola opens upwards, so the expression is non-negative when x -2 or x 1. 2. The denominator cannot be zero: x-1 0 x 1. Combining these conditions: x -2 or x 1, and x 1. This means x -2 or x > 1. The domain is D_f_7 = (-, -2] (1, +).