value of x at which the slope of the tangent curve defined by y=(2x+1)^(1/3)(x^2-x-6) is horizontal?

Answer 1

By the product rule,

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

Now, factor out (2x+1)^(-2/3). You then get

y' = (2x+1)^(-2/3)[(2/3)(x^2-x-6) + (2x-1)(2x+1)]
= (2x+1)^(-2/3)[(2/3)x^2 - (2/3)x - 4 + 4x^2 - 1]
= (2x+1)^(-2/3)[(14/3)x^2 - (2/3)x - 5]
= (1/3)(2x+1)^(-2/3)[14x^2 - 2x - 15].

A horizontal tangent occurs when y' = 0, which will occur when 14x^2 - 2x - 15 = 0. From the quadratic formula, the roots are

x = (2 +/- sqrt(844))/28 = (1 +/- sqrt(211))/14,

so these are the x-values at which the tangent is horizontal.

Answer 2

The slope of the tangent curve is given by :
y'=(1/3)(2x+1)^(-2/3).2. (x^2-x-6) + (2x+1)^(1/3)(2X-1)

We need to determine when y' equals 0.

=> (2/3)(x^2-x-6) + (2x+1)(2x-1) = 0
=> (2x^2 - 2x - 12) + 3(4x^2 - 1) = 0
=> (2x^2 - 2x - 12 + 12x^2 - 3) = 0
=> (14x^2 - 2x - 15) = 0

=> x = (2 + (844)^(1/2))/28
OR
x = (2 - (844)^(1/2))/28