WebWe have, f (x) = 3 x 2 + 6 x + 8 ⇒ f (x) = 3 (x 2 + 2 x + 1) + 5 = 3 (x + 1) 2 + 5. Now, 3 (x + 1) 2 ≥ 0 for all x ∈ R ⇒ 3 (x + 1) 2 + 5 ≥ 5 for all x ∈ R ⇒ f (x) ≥ f (− 1) for all x ∈ R. Thus, 5 is the minimum value of f (x) which it attains at x = − 1. Since, f (x) can be made as large as we please. Therefore, the ... Web(- 3x 2 + 3x + 4) Concept used: dy/dx = 0, the value of x gives the minimum value when d 2 y/dx 2 is greater than 0, and gives the maximum value when d 2 y/dx 2 is less than 0 Calculation: (- 3x 2 + 3x + 4) ---- (i) Differentiating (i), dy/dx = 0 ⇒ - 6x + 3 = 0 ---- (ii) ⇒ x = 1/2 By double differentiating equation (ii), d 2 y/dx 2 = - 6 < 0
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WebFirst of all, distribute the fraction. (3x^2+9x+17)/ (3x^2+9x+7) = 1 + 10/ (3x^2+9x+7) Now, for this sum to be Maximum, the fraction should be at its maximum value. For this to happen, the denominator of the fraction should be at its minimum. Consider the polynomial 3x^2+9x+7. Try to make the polynomial a Perfect Square. WebCorrect option is C) y= x 2+2x+3x 2+14x+9=1+ x 2+2x+312x+6 y= (x 2+2x+3) 2(x 2+2x+3)(12)−(12x+6)(2x+2) y= (x 2+2x+3) 2−12x 2−12x+24. for max-min, y=0. so, x … oxford physical therapy cincinnati oh
If x is real the maximum value of 3x^2+9x+17/3x^2+9x+7 is
Web27 mrt. 2024 · answered Mar 27, 2024 by Niharika (75.9k points) selected Mar 28, 2024 by Vikash Kumar. Best answer. ⇒ 2yx2 + 3yx + 6y = x + 2. ⇒ 2yx2 + (3y – 1) x + 6y – 2 = 0. … WebCalculation: ⇒ Let y = \(\dfrac{(x^2-2x+4)}{(x^2+2x+4)}\) \(⇒ y(x^2+2x+4) = x^2-2x+4)\) \(⇒x^2 (y-1)+x(2y+2)+4y-4=0 )\) Since x is real, the discriminant of the ... Web14 okt. 2024 · If x is real, then maximum value of (3x2 + 9x + 17)/ (3x2 + 9x + 7) is (a) 41 (b) 1 (c) 17/7 (d) 1/4 limits continuity and differntiability jee jee mains 1 Answer +1 vote answered Oct 14, 2024 by KajalAgarwal (44.7k points) selected Oct 16, 2024 by Vikash Kumar Best answer Correct Option (a) 41 Explanation: ← Prev Question Next Question → oxford physical therapy maineville ohio