Expert: Ahmed Salami - 3/8/2006

Question
Find the points on the graph of y = (x^3)/3 - 5x - 4/x at which the slope of the tangent is horizontal.

Answer
Hi Dave,
Sorry for the time it took.
The tangent of a curve at any point is horizontal when
dy/dx = 0 (i.e when the curve is at a turning point,
minimum or maximum)
For y = x^3/3 - 5x - 4/x
y = (1/3)x^3 - 5x^1 - 4(x^-1)
Differentiating y = ax^n gives us dy/dx = nax^(n-1)
And so
dy/dx = 3(1/3)x^2 - (1)5x^0 - (-1)4(x^-2)
     = x^2 - 5 + 4(x^-2)
     = x^2 - 5 + 4/x^2
The tangent would be horizontal when
x^2 - 5 + 4/x^2 = 0
Multiplying through by x^2
x^4 - 5x^2 + 4 = 0
This is a quadratic equation in x^2, and can be solved
by taking, say p = x^2. We arrive at
p^2 - 5p + 4 = 0
Which is now a quadratic equation in p, solvable by
factorisation.
(p-4)(p-1) = 0
p = 4 or 1
i.e
x^2 = 4      OR     x^2 = 1
x = +or- 2   OR     x = +or- 1
Which gives the four values of x at which the tangents
to the curve are horizontal.
At x = -1, y = (-1^3)/3 - 5(-1) - 4/(-1)
            = -1/3 + 5 + 4 = 26/3
At x = 1, y = (1^3)/3 - 5(1) - 4/(1)
           = 1/3 - 5 - 4 = -26/3
At x = -2, y = (-2^3)/3 - 5(-2) - 4/(-2)
            = -8/3 + 10 + 2 = 28/3
At x = 2, y = (2^3)/3 - 5(2) - 4/(2)
           = 8/3 - 10 - 2 = -28/3
The four points are;
(-1,26/3),(1,-26/3),(-2,28/3),(2,-28/3)

I hope i have helped. You can always get back to me.
Regards.