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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calendarEducator since 2008
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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)
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calendarEducator since 2010
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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 = 1, y = (1/3) - 5 - 4 = -26/3
For x = -1, y = -1/3 + 5 + 4 = 26/3
For x = 2, y = 8/3 - 2*5 - 4/2 = -28/3
For x = -2, y = (-8/3) + 4/2 + 2*5 = 28/3
The points at which the tangent is horizontal are:
(1 , -26/3) , ( -1, 26/3) , (2, -28/3) and ( -2, 28/3)
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