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...

First of all, finding the range of a function involves finding all possible values that the function can take. It is not necessary to take the derivative and find critical values to determine the range of a rational function like y = (3x-1)/(2x^2 + x - 6).

To find the range, it is helpful to analyze the behavior of the function as x approaches positive infinity and negative infinity.

As x approaches positive infinity, both the numerator and denominator of the function go to infinity. Since the numerator has a higher power of x than the denominator, the function approaches positive infinity as x approaches positive infinity.

As x approaches negative infinity, both the numerator and denominator of the function also go to infinity. Again, since the numerator has a higher power of x than the denominator, the function approaches positive infinity as x approaches negative infinity.

So, we can conclude that the range of the function y = (3x-1)/(2x^2 + x - 6) is all real numbers except for zero. In other words, the range is (-∞, 0) U (0, +∞).

Now, regarding the issue you encountered with imaginary numbers while taking the derivative, it seems that you made an error in your calculations. When finding the derivative of the function, there should not be any imaginary numbers involved. However, it is not necessary to find the derivative to determine the range in this case.

To summarize:

1. The range of y = (3x-1)/(2x^2 + x - 6) is (-∞, 0) U (0, +∞).
2. You did not need to take the derivative to find the range.
3. The appearance of imaginary numbers in your calculations might have been due to a mistake in your calculations. Double-check your work to ensure accuracy.