Deux problèmes sont présentés: le premier porte sur les transformations du plan, et le second sur l'étude d'une fonction et le calcul d'une aire.

Problème 2, Partie A: 1) Déterminer la dérivée de f sur [0; +[. Step 1: Identify the function and its components for differentiation. The function is f(x) = ((1+x))^2. This is a composite function of the form u^2, where u(x) = (1+x). The derivative of u(x) is u'(x) = (d)/(dx)((1+x)) = (1)/(1+x). Step 2: Apply the chain rule (u^n)' = n u^n-1 u'. f'(x) = 2 · (1+x) · (1)/(1+x) f'(x) = (2(1+x))/(1+x) The derivative is defined for 1+x > 0, which means x > -1. Thus, it is defined on [0; +[. The derivative of f on [0; +[ is (2(1+x))/(1+x). 2) En déduire les variations de f sur [0; +[. Step 1: Analyze the sign of the denominator. For x [0; +[, we have x 0, so 1+x 1. Therefore, 1+x > 0. Step 2: Analyze the sign of the numerator. For x [0; +[, we have 1+x 1. Since the natural logarithm function (t) is non-negative for t 1, it follows that (1+x) (1) = 0. Therefore, 2(1+x) 0. Step 3: Conclude the sign of f'(x) and the variations of f(x). Since 2(1+x) 0 and 1+x > 0 for x [0; +[, we have f'(x) = (2(1+x))/(1+x) 0. f'(x) = 0 if and only if (1+x) = 0, which implies 1+x = 1, so x=0. As f'(x) 0 on [0; +[, the function f(x) is increasing on this interval. 3) Calculer f(0) et f(1), puis dresser le tableau de variations de f sur [0; 1]. Step 1: Calculate f(0). f(0) = ((1+0))^2 = ((1))^2 = 0^2 = 0 Step 2: Calculate f(1). f(1) = ((1+1))^2 = ((2))^2 Step 3: Construct the table of variations for f on [0; 1]. From part 2, f'(x) 0 on [0; +[, so f'(x) 0 on [0; 1]. The function f is increasing on [0; 1]. The table of variations for f on [0; 1] is: |c|ccc| x & 0 & & 1 \\ f'(x) & 0 & + & \\ f(x) & 0 & & ((2))^2 \\ The values are f(0) = 0 and f(1) = ((2))^2. 4) Calculer f'(0). En déduire l'équation de la tangente T à C au point d'abscisse 0. Step 1: Calculate f'(0). Using the derivative f'(x) = (2(1+x))/(1+x) from part 1: f'(0) = (2(1+0))/(1+0) = (2(1))/(1) = (2 · 0)/(1) = 0 So, f'(0) = 0. Step 2: Determine the equation of the tangent T. The equation of the tangent line at a point (x_0, f(x_0)) is given by y - f(x_0) = f'(x_0)(x - x_0). Here, x_0 = 0. We have f(0) = 0 (from part 3) and f'(0) = 0. y - 0 = 0(x - 0) y = 0 The equation of the tangent T is y=0. 5) Tracer dans (o; i, j) C en faisant apparaître la tangente T. This step requires a graphical representation. Here are the key characteristics for drawing the graph: • The curve C passes through the origin (0,0). • The tangent T at the origin is the x-axis (y=0). This means the curve is tangent to the x-axis at (0,0). • The function f(x) is increasing on [0; +[. • The point (1, f(1)) = (1, ((2))^2) is on the curve. Approximately, ((2))^2 ≈ (0.693)^2 ≈ 0.48. • The graphical unit is 10 cm for 1 unit. So, the point (1, 0.48) would be 10 cm along the x-axis and 4.8 cm along the y-axis from the origin. • As x increases, f(x) increases without bound. What's next? Send 'em! 📸