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

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We have to find the point on the graph where the tangent is horizontal.

When a line is horizontal its slope is zero. The slope of the tangent to a graph at any point can be found by finding the value of the first derivative at that point.

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

y' = (1/3)*3*x^2 - 5 + 4/x^2

Equate y' to zero

(1/3)*3*x^2 - 5 + 4/x^2 = 0

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

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

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

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

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

This gives x^2 = 1 and x^2 = 4

x = -1 and + 1 and x = 2 and -2

For x =...

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