Calc I
posted by Kate .
Find the range of
y = (3x1)/(2x^2 + x  6)
I was going to take the derivative of the numerator and denomenator and then use the quotent rule to find the derivative of the function and find the critical values but I ran into the imaginary number during the proess and am woundering if I did nothing wrong... if I did nothing wrong then what exactly does this mean?
y = (3x1)/(2x^2 + x  6) = h(x)
f(x) = (3x1)
g(x)= 2x^2 + x  6
dy/dx f(x) = ( 3(x + a) 1 3x + 1 ) / a = (3x + 3a 1  3x +1)/a = (3a)/a = 3
dy/dx f(g) = ( 2(x + a)^2 + x + a  6 2x^2 x +6)/a = (2(x^2 + a^2 + 2ax) + a 2x^2)/a = (2x^2 + 2a^2 + 4ax + a 2x^2)/a = (2a^2 + 4ax +a)/a = 2a + 4x + 1 = 4x + 1
dy/dx h(x) = ( (2x^2 + x 6)3(3x  1)(4x +1) )/ (2x^2 + x  6)^2 = ( 6x^2 + 3x  18 (12x^2 + 3x  4x 1) )/(2x^2 + x  6)^2 = ( 6x^2 + 3x  18 (12x^2 x  1) )/(2x^2 + x  6)^2 = ( 6x^2 + 3x  18 12x^2 + x + 1)/(2x^2 + x  6)^2 = ( 6x^2 + 4x 17)/(2x^2 + x  6)^2
At this point I proceded by setting the numerator to zero and solving which is were I got stuck at least I think I don't see what I have done any where even while typing this thing up charecter by charecter, normally this is a good way to catch mistakes but I have found nothing wrong with my work...
6x^2 + 4x 17 = 0
(4 +/ sqrt( 4^2  4(6)(17) ) )/(2(6)) = 4/12 +/ (1/12)sqrt( 4  (6)(17) ) = 1/3 +/ (1/6)sqrt(2(2  (3)(17)) = 1/3 +/ (1/6)sqrt(2(251)) = 1/3 +/ (1/6)sqrt(2(47) = 1/3 +/ (7/6)sqrt(2) = 1/3 +/ (7i/6)sqrt(2)
Is this correct or did I just simply do something wrong here? Don't see were or how...
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