On considère la fonction f définie sur IR par : f(x) = ex/(e^x+1) + 2. On note par (C) sa courbe représentative dans un repère orthonormé (O,i,j) d'unité 2cm.

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Asked on July 14, 2026Mathematics

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.

Accepted Answer · 2026-07-14T07:49:47.873Z

Step 1: Calculate _x - f(x) and interpret graphically. The function is f(x) = (e^x)/(e^x+1) + 2. As x -, e^x 0. _x - f(x) = _x - ((e^x)/(e^x+1) + 2) = (0)/(0+1) + 2 = 0 + 2 = 2 Graphically, this means the line y=2 is a horizontal asymptote to the curve (C) as x -. Step 2: Demonstrate f(x) = 3 - (1)/(e^x+1) and deduce _x + f(x). a) We start with the given expression for f(x): f(x) = (e^x)/(e^x+1) + 2 We can rewrite the fraction (e^x)/(e^x+1) as (e^x+1-1)/(e^x+1) = 1 - (1)/(e^x+1). Substitute this back into the expression for f(x): f(x) = (1 - (1)/(e^x+1)) + 2 = 3 - (1)/(e^x+1) b) Now, we deduce _x + f(x) using the new form: As x +, e^x +. Therefore, e^x+1 +, which implies (1)/(e^x+1) 0. _x + f(x) = _x + (3 - (1)/(e^x+1)) = 3 - 0 = 3 Graphically, this means the line y=3 is a horizontal asymptote to the curve (C) as x +. Step 3: Verify f'(x) = (e^x)/((e^x+1)^2) and create the variation table. a) We differentiate f(x) = (e^x)/(e^x+1) + 2. The derivative of the constant 2 is 0. Let u(x) = e^x and v(x) = e^x+1. Then u'(x) = e^x and v'(x) = e^x. Using the quotient rule ((u)/(v))' = (u'v - uv')/(v^2): f'(x) = (e^x(e^x+1) - e^x(e^x))/((e^x+1)^2) = e^2x + e^x - e^2x(e^x+1)^2 = (e^x)/((e^x+1)^2) This matches the given derivative. b) To create the variation table, we analyze the sign of f'(x). For all real x, e^x > 0. For all real x, (e^x+1)^2 > 0. Therefore, f'(x) > 0 for all real x. This means f(x) is strictly increasing on R. The limits are _x - f(x) = 2 and _x + f(x) = 3. The table of variation for f is: |c|ccc| x & - & & + \\ f'(x) & & + & \\ f(x) & 2 & & 3 \\ Step 4: Show I(0; (5)/(2)) is a center of symmetry, find the tangent equation, and calculate function values. a) A point (a,b) is a center of symmetry for f(x) if f(a-x) + f(a+x) = 2b. Here, (a,b) = (0, (5)/(2)), so we need to show f(-x) + f(x) = 2 × (5)/(2) = 5. We use f(x) = (e^x)/(e^x+1) + 2. First, find f(-x): f(-x) = e^-xe^-x+1 + 2 = (1)/(e^x)(1)/(e^x)+1 + 2 = (1)/(e^x)(1+e^x)/(e^x) + 2 = (1)/(e^x+1) + 2 Now, add f(x) and f(-x): f(-x) + f(x) = ((1)/(e^x+1) + 2) + ((e^x)/(e^x+1) + 2) = (1+e^x)/(e^x+1) + 4 = 1 + 4 = 5 Since f(-x) + f(x) = 5, the point I(0; (5)/(2)) is a center of symmetry for the curve (C). b) The equation of the tangent (T) at x_0=0 is y - f(x_0) = f'(x_0)(x - x_0). First, calculate f(0): f(0) = (e^0)/(e^0+1) + 2 = (1)/(1+1) + 2 = (1)/(2) + 2 = (5)/(2) Next, calculate f'(0): f'(0) = (e^0)/((e^0+1)^2) = (1)/((1+1)^2) = (1)/(2^2) = (1)/(4) Substitute these values into the tangent equation: y - (5)/(2) = (1)/(4)(x - 0) y = (1)/(4)x + (5)/(2) The equation of the tangent (T) is y = (1)/(4)x + (5)/(2). c) Calculate f(-1), f(1), and f(2) to 0.1 precision using e^-1 ≈ 0.4, e ≈ 2.7, e^2 ≈ 7.4. For f(-1): f(-1) = e^-1e^-1+1 + 2 ≈ (0.4)/(0.4+1) + 2 = (0.4)/(1.4) + 2 ≈ 0.2857 + 2 = 2.2857 ≈ 2.3 For f(1): f(1) = (e^1)/(e^1+1) + 2 ≈ (2.7)/(2.7+1) + 2 = (2.7)/(3.7) + 2 ≈ 0.7297 + 2 = 2.7297 ≈ 2.7 For f(2): f(2) = (e^2)/(e^2+1) + 2 ≈ (7.4)/(7.4+1) + 2 = (7.4)/(8.4) + 2 ≈ 0.8809 + 2 = 2.8809 ≈ 2.9 Step 5: Trace the curve (C) and (T) in the same coordinate system. This step requires a graphical representation. Key elements for tracing are: • Horizontal asymptotes: y=2 (as x -) and y=3 (as x +). • Center of symmetry: I(0, 2.5). • Tangent line (T): y = (1)/(4)x + 2.5, which passes through I(0, 2.5) with a slope of (1)/(4). • Points: f(-1) ≈ 2.3, f(0) = 2.5, f(1) ≈ 2.7, f(2) ≈ 2.9. • The curve (C) is strictly increasing, approaching y=2 from above and y=3 from below. Step 6: Calculate F'(x) and the area A. a) Let F(x) = 2x + (e^x+1). To calculate F'(x), we differentiate each term: F'(x) = (d)/(dx)(2x) + (d)/(dx)((e^x+1)) F'(x) = 2 + (e^x)/(e^x+1) We observe that F'(x) = f(x). Conclusion: F(x) is an antiderivative (or primitive) of f(x). b) We need to calculate the area A bounded by the curve (C), the x-axis, and the lines x=0 and x= 3. Since f(x) > 2 for all x, the curve is always above the x-axis. The area is given by the definite integral: A_units = _0^ 3 f(x)\,dx Since F(x) is an antiderivative of f(x): A_units = [F(x)]_0^ 3 = F( 3) - F(0) Calculate F( 3): F( 3) = 2( 3) + (e^ 3+1) = 2 3 + (3+1) = 2 3 + 4 = 2 3 + 2 2 Calculate F(0): F(0) = 2(0) + (e^0+1) = 0 + (1+1) = 2 Now, substitute these values: A_units = (2 3 + 2 2) - 2 = 2 3 + 2 Using the given approximations 2 ≈ 0.7 and 3 ≈ 1.1: A_units ≈ 2(1.1) + 0.7 = 2.2 + 0.7 = 2.9 The problem states that the unit on the axes is 2 cm. This means 1 unit of area in the coordinate system corresponds to (2 cm) × (2 cm) = 4 cm^2. Therefore, the area in cm^2 is: A = A_units × 4 cm^2 ≈ 2.9 × 4 = 11.6 cm^2 The geometric area is 11.6 cm^2. Send me the next one 📸

Accepted Answer · 2026-07-14T07:49:47.873Z

Step 1: Calculate _x - f(x) and interpret graphically. The function is f(x) = (e^x)/(e^x+1) + 2. As x -, e^x 0. _x - f(x) = _x - ((e^x)/(e^x+1) + 2) = (0)/(0+1) + 2 = 0 + 2 = 2 Graphically, this means the line y=2 is a horizontal asymptote to the curve (C) as x -. Step 2: Demonstrate f(x) = 3 - (1)/(e^x+1) and deduce _x + f(x). a) We start with the given expression for f(x): f(x) = (e^x)/(e^x+1) + 2 We can rewrite the fraction (e^x)/(e^x+1) as (e^x+1-1)/(e^x+1) = 1 - (1)/(e^x+1). Substitute this back into the expression for f(x): f(x) = (1 - (1)/(e^x+1)) + 2 = 3 - (1)/(e^x+1) b) Now, we deduce _x + f(x) using the new form: As x +, e^x +. Therefore, e^x+1 +, which implies (1)/(e^x+1) 0. _x + f(x) = _x + (3 - (1)/(e^x+1)) = 3 - 0 = 3 Graphically, this means the line y=3 is a horizontal asymptote to the curve (C) as x +. Step 3: Verify f'(x) = (e^x)/((e^x+1)^2) and create the variation table. a) We differentiate f(x) = (e^x)/(e^x+1) + 2. The derivative of the constant 2 is 0. Let u(x) = e^x and v(x) = e^x+1. Then u'(x) = e^x and v'(x) = e^x. Using the quotient rule ((u)/(v))' = (u'v - uv')/(v^2): f'(x) = (e^x(e^x+1) - e^x(e^x))/((e^x+1)^2) = e^2x + e^x - e^2x(e^x+1)^2 = (e^x)/((e^x+1)^2) This matches the given derivative. b) To create the variation table, we analyze the sign of f'(x). For all real x, e^x > 0. For all real x, (e^x+1)^2 > 0. Therefore, f'(x) > 0 for all real x. This means f(x) is strictly increasing on R. The limits are _x - f(x) = 2 and _x + f(x) = 3. The table of variation for f is: |c|ccc| x & - & & + \\ f'(x) & & + & \\ f(x) & 2 & & 3 \\ Step 4: Show I(0; (5)/(2)) is a center of symmetry, find the tangent equation, and calculate function values. a) A point (a,b) is a center of symmetry for f(x) if f(a-x) + f(a+x) = 2b. Here, (a,b) = (0, (5)/(2)), so we need to show f(-x) + f(x) = 2 × (5)/(2) = 5. We use f(x) = (e^x)/(e^x+1) + 2. First, find f(-x): f(-x) = e^-xe^-x+1 + 2 = (1)/(e^x)(1)/(e^x)+1 + 2 = (1)/(e^x)(1+e^x)/(e^x) + 2 = (1)/(e^x+1) + 2 Now, add f(x) and f(-x): f(-x) + f(x) = ((1)/(e^x+1) + 2) + ((e^x)/(e^x+1) + 2) = (1+e^x)/(e^x+1) + 4 = 1 + 4 = 5 Since f(-x) + f(x) = 5, the point I(0; (5)/(2)) is a center of symmetry for the curve (C). b) The equation of the tangent (T) at x_0=0 is y - f(x_0) = f'(x_0)(x - x_0). First, calculate f(0): f(0) = (e^0)/(e^0+1) + 2 = (1)/(1+1) + 2 = (1)/(2) + 2 = (5)/(2) Next, calculate f'(0): f'(0) = (e^0)/((e^0+1)^2) = (1)/((1+1)^2) = (1)/(2^2) = (1)/(4) Substitute these values into the tangent equation: y - (5)/(2) = (1)/(4)(x - 0) y = (1)/(4)x + (5)/(2) The equation of the tangent (T) is y = (1)/(4)x + (5)/(2). c) Calculate f(-1), f(1), and f(2) to 0.1 precision using e^-1 ≈ 0.4, e ≈ 2.7, e^2 ≈ 7.4. For f(-1): f(-1) = e^-1e^-1+1 + 2 ≈ (0.4)/(0.4+1) + 2 = (0.4)/(1.4) + 2 ≈ 0.2857 + 2 = 2.2857 ≈ 2.3 For f(1): f(1) = (e^1)/(e^1+1) + 2 ≈ (2.7)/(2.7+1) + 2 = (2.7)/(3.7) + 2 ≈ 0.7297 + 2 = 2.7297 ≈ 2.7 For f(2): f(2) = (e^2)/(e^2+1) + 2 ≈ (7.4)/(7.4+1) + 2 = (7.4)/(8.4) + 2 ≈ 0.8809 + 2 = 2.8809 ≈ 2.9 Step 5: Trace the curve (C) and (T) in the same coordinate system. This step requires a graphical representation. Key elements for tracing are: • Horizontal asymptotes: y=2 (as x -) and y=3 (as x +). • Center of symmetry: I(0, 2.5). • Tangent line (T): y = (1)/(4)x + 2.5, which passes through I(0, 2.5) with a slope of (1)/(4). • Points: f(-1) ≈ 2.3, f(0) = 2.5, f(1) ≈ 2.7, f(2) ≈ 2.9. • The curve (C) is strictly increasing, approaching y=2 from above and y=3 from below. Step 6: Calculate F'(x) and the area A. a) Let F(x) = 2x + (e^x+1). To calculate F'(x), we differentiate each term: F'(x) = (d)/(dx)(2x) + (d)/(dx)((e^x+1)) F'(x) = 2 + (e^x)/(e^x+1) We observe that F'(x) = f(x). Conclusion: F(x) is an antiderivative (or primitive) of f(x). b) We need to calculate the area A bounded by the curve (C), the x-axis, and the lines x=0 and x= 3. Since f(x) > 2 for all x, the curve is always above the x-axis. The area is given by the definite integral: A_units = _0^ 3 f(x)\,dx Since F(x) is an antiderivative of f(x): A_units = [F(x)]_0^ 3 = F( 3) - F(0) Calculate F( 3): F( 3) = 2( 3) + (e^ 3+1) = 2 3 + (3+1) = 2 3 + 4 = 2 3 + 2 2 Calculate F(0): F(0) = 2(0) + (e^0+1) = 0 + (1+1) = 2 Now, substitute these values: A_units = (2 3 + 2 2) - 2 = 2 3 + 2 Using the given approximations 2 ≈ 0.7 and 3 ≈ 1.1: A_units ≈ 2(1.1) + 0.7 = 2.2 + 0.7 = 2.9 The problem states that the unit on the axes is 2 cm. This means 1 unit of area in the coordinate system corresponds to (2 cm) × (2 cm) = 4 cm^2. Therefore, the area in cm^2 is: A = A_units × 4 cm^2 ≈ 2.9 × 4 = 11.6 cm^2 The geometric area is 11.6 cm^2. Send me the next one 📸

On considère la fonction f définie sur IR par : f(x) = ex/(e^x+1) + 2. On note par (C) sa courbe représentative dans un repère orthonormé (O,i,j) d'unité 2cm.

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On considère la fonction f définie sur IR par : f(x) = ex/(e^x+1) + 2. On note par (C) sa courbe représentative dans un repère orthonormé (O,i,j) d'unité 2cm.

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