Find the range of
y = (3x-1)/(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 = (3x-1)/(2x^2 + x - 6) = h(x)
f(x) = (3x-1)
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(2-51)) = 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...
3 answers