if 'x' is real ,the maximum value of 3x^2 + 9x = 17 /3x^2 +9x + 7 is

find the minimum value
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1 Answer

(3x2+9x+17)/(3x2+9x+7)=1+10/(3x2+9x+7).

When x is very large (positive or negative), the expression approaches 1 because 10/(3x2+9x+7)→0. So 1 is an asymptote.

3x2+9x+7 is never zero for real x, but the maximum value of the expression is when 10/(3x2+9x+7) is maximum: when the derivative is zero:

-10(3x2+9x+7)-2)(6x+9)=0, so 6x=-9, x=-3/2 or -1.5.

10/(3x2+9x+7)=10/(6.75-13.5+7)=10/0.25=40. So the maximum is 40+1=41.

by Top Rated User (1.1m points)

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