Find the points on the graph of y= 1/3 x^3 - 5x - 4/x at which the tangent is horizontal.

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

We need to find the horizontal tangent line to the
curve y.

==> The equation of the line
is:

y-y1 = m(x-x1) where m is the
slope.

But we know that the horizontal line has a slope of
zero.

==> y= y1 ( y1 is the point of tendency ( 0,
y1)

Now we will differentiate the curve
y.

==> y' = x^2 - 5 +
4/x^2

Now we will find the point of tendency where the
slope is zero.

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

==> We will multiply by
x^2

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

Now we will
factor.

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

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

Then the point where the curve has tangent line
are:

x1= 2 ==> y1= 8/3 - 10 - 2 =
-28/3

x2= -2  ==> y2 = -8/3 +10 +2 =
28/3

x3= 1==> y3= 1/3 -5 -4 =
-26/3

x4= -1 ==> y4 = -1/3 + 5 +4 =
26/3

Then the points where the tangents are
horizontal are:

(2,-28/3) ,
(-2, 28/3), (1, -26/3) , and (-1, 26/3)