Find the range of

Question

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...

SO I put the equation into a derivative calculator and got the same results so there are no turning points? I'm sure that there has to be on no? Thye are there they are just complex correct?
16 minutes ago

Using the calculate max on the graphing calculator I get a turning point at x = .999999894 with a local max of 1/8

so the range is (1/8, infinity) now how do I do this without a graphing calculator and cheating setting the derivitive to zero gave me imaginary number... please help me as I even put the funciton into a derivitive calculator and got the same derivitive I found... What do I do
7 minutes ago

Your wrong y equals zero when x=1/3 and no the equation has y values greater than y=0 I found the turning point see above
6 minutes ago

sorry should be so the range is (1/8, -infinity)

Answer 1

Your derivative of the function using the quotient rule is correct.
setting that equal to zero to find turning points and only getting imaginary roots is also correct.
Thus there are not turning points, or there are not max and minimum points.

looking at the denominator, which factors to (2x-3)(x+2) we can see that there are two vertical asymptotes, one at x=3/2, the other at x = 2.

So the range would be from -infinity to + infinity.

If we let the numerator of 3x-1 equal to zero we get x = 1/3, so there is an x-intercept of 1/3

When x --> - infinity, y ---> 0 from below the x-axis, when x ---> +infinity , y --->0 from above the x-axis

I see a graph made up of 3 distinct parts.
One part below the x-axis and to the left of the vertical asymptote x = -2
Another part between the asymptotes x = -2 and x = 3/2 starting at the top and dropping to the bottom, crossing the x=axis at x=1/3
Third part above dropping from x=3/2 and approaching the x-axis.

Just like in yesterday's question, I do not understand why you are finding the derivatives of numerator and denominator in separate parts.
Not only that, but you appear to be using some strange method which resembles finding the derivative using First Principles.

Answer 2

the 6th line above should have said

"vertical asymptotes, one at x=3/2, the other at x = - 2. "