Expert: Paul Klarreich - 3/8/2006

Question
Fine 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,
You wrote:
Subject:  Slope of a Tangent
Question:  Fine the points on the graph of y = (x^3)/3 - 5x - 4/x at

which the slope of the tangent is horizontal
------------------------------
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4. Indicate what you already tried to do and, if possible, where you got stuck.

For this question:

y = (x^3)/3 - 5x - 4/x
y = (x^3)/3 - 5x - 4x^-1

dy
-- = x^2 - 5 + 4x^-2
dx

dy              4
-- = x^2 - 5 + ---
dx             x^2

Now about the slope of the tangent.  Your phrasing is bad.  That may be why you are having trouble.  You should be saying EITHER:
The tangent is horizontal.
OR
The slope of the tangent is zero.

The slope of the tangent is a number.  The only way numbers can be horizontal is to write them on a sheet of paper and lie it flat on the floor.

So you want the slope to be zero.  The slope of the tangent is the value of the derivative, so you set dy/dx = 0 and solve:

           4
x^2 - 5 + --- = 0
          x^2

Multiply through by x^2

x^4 - 5x^2 + 4 = 0

Factor:

(x^2 - 4)(x^2 - 1) = 0

(x + 2)(x - 2)(x + 1)(x - 1) = 0

x = +- 2  [Substitute into y = ... to get the y-coordinate.]
x = +- 1  [ ...............................................]

I'll leave that part to you.