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If x is real,the maximum value of 3x 2 +9x+17/(3x 2 +9x+7) is

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3 years ago

```								First 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. This way, it'll be easier to calculate its Minimum.					3x^2+9x+7 = [{(√3)x}^2 + 2*(√3)*{3(√3)/2}*x + 27/4] +(7 - 27/4)			This is nothing but, [(√3)x + 3*(√3)/2]^2 + 1/4.			This polynomial can have its minimum value as 1/4, because the Perfect Square can have its minimum value as 0, since x is Real.						Thus min value of polynomial is 1/4.		Hence total value(maximum) = 1+ 10/(1/4)		= 1+ 40 = 41
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3 years ago 605 Points
```							Dear student,Please find the solution to your problem below. Let, T = 3x2+9x+17/(3x2+9x+7) = 1 + 10/(3x2+9x+7)Now, for its max value (3x2+9x+7) needs to be minimum.Now, 3x2+9x+7 = 3(x2+3x+7/3)                       = 3[(x2 + 2(3/2)x + 9/4) + (7/3 – 9/4)]                       = 3(x+3/2)2 + ¼Hence, min.(3x2+9x+7) = ¼ Hence Tmax = 1 + 10/(1/4) = 1 + 40 = 41 Thanks and regards,Kushagra
```
2 months ago
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