Help with derivative & points on curve

[math_student03]

Hey guys, really stuck on this problem, thanks for the help !

Question:
Find the values of x at which the slope of the tangent line to the curve defined by y=(x+1)1/3(x2-x-6) is:
a) Undefined / vertical b) horizontal

Answer / Attempted answer

Firstly I took the derivative:

Dy/dx = 1/3 (2x+1)^-2/3(2)(x2-x-6) + (2x-1)(2x+1) ^1/3
Dy/dx = (2x+1) ^-2/3 [ 2/3(x-3)(x+2) + (2x-1)(2x+1) ^-1/2]


Now here I am stuck, I don’t know how to get part a) and for part b) would I just set dy/dx to zero and solve for the x values?

 

[danago]

Well when is a fraction undefined?

 

[math_student03]

when the denom = 0?

.. but there isn't one, or should i transfer the - exponents to the botton and create one?

 

[danago]

(x+1)1/3

is that (x+1)^1/3?

Or is it the product of (x+1) and 1/3?

 

[HallsofIvy]

Hey guys, really stuck on this problem, thanks for the help !

Question:
Find the values of x at which the slope of the tangent line to the curve defined by y=(x+1)1/3(x2-x-6) is:
a) Undefined / vertical b) horizontal

Peculiarly stated! If the slope is "undefined", then the line itself is vertical. If the slope is 0, then the line itself is horizontal.
I assume that the function given is y= (x+1)/(3(x^2- x-6)) is that correct?
What is that extra "1" in the numerator? The other possible interpretation of what you wrote is y= (x+1)(1/3)(x^2- x- 6) but that seems unlikely. From what you write further, perhaps that "1/3(x2-x-6)" is ^3\sqrt{x^2- x-6}= (x+1)(x^2-x-6)^{1/3} but if so that really bad notation!

Answer / Attempted answer

Firstly I took the derivative:

Dy/dx = 1/3 (2x+1)^-2/3(2)(x2-x-6) + (2x-1)(2x+1) ^1/3
Dy/dx = (2x+1) ^-2/3 [ 2/3(x-3)(x+2) + (2x-1)(2x+1) ^-1/2]

Assuming y= (x+1) (x^2-x-6)^{1/3} then dy/dx= (x^2-x-6)^{1/3}+ (1/3)(x+1)(x^2-x-6)^{-2/3}(2x-1)

Now here I am stuck, I don’t know how to get part a) and for part b) would I just set dy/dx to zero and solve for the x values?

That y' certainly does have a "denominator" because of that -2/3 power. For what value of x is that denominator = 0? And, for b, yes, you set the whole thing equal to 0 and solve for x.

 

[math_student03]

ok the original function was:

y= (x+1)^{1/3} (x^2-x-6)

i then got
dy/dx= (1/3)(2x+1)^{-2/3}(2)(x^2-x-6) + (2x-1)(2x+1)^{1/3}

now here is my dy/dx , from here would i common factor out the (2x+1)^{-2/3} or just put it right to the bottom and get the demon. on one part and set that to 0 to see where it is undefined. (assuming this it would be definied at x=-1/2)