# find the points on the graph of y=1/3 x^3-5 x-4/x at which the slope of the tangent is horizontal how do you use the first principle to solve the questio y= (1/3) x^3 - 5x - 4/x

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

==> The horizontal tangent has a lope of 0. Then we need to find the point where the slope y' is 0.

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

Now we will multiply by...

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

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

==> The horizontal tangent has a lope of 0. Then we need to find the point where the slope y' is 0.

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

Now we will multiply by x^2

=> x^4 -5x^2 + \$ = 0

Now we will factor.

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

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

==> x = 2, -2, 1, -1

Then, the points where the curve y has a tangent line are the following:

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

x2= -2 ==> y2 = 28/3

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

x4 = -1 ==> y4= 26/3

Then the points are :

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

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