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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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)

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial