Find the points on the graph of y=(1/3)x^3 - 5x - 4/x where the tangent is horizontal

[amak]

Hi there,

I am given the following problem, this is in the section of the textbook about difference quotient to find slope.

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22. Find the points in the graph of the following function:

. . . . .\(\displaystyle y\, =\, \dfrac{1}{3}x^3\, -\, 5x\, -\, \dfrac{4}{x}\)

...at which the tangent is horizontal.

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I have not learned derivatives yet. I'm thinking I'm doing something with the fact that the slope of the tangent in this case is 0?

Thanks in advance!

 

[Deleted member 4993]

Hi there,

I am given the following problem, this is in the section of the textbook about difference quotient to find slope.

---

22. Find the points in the graph of the following function:

. . . . .\(\displaystyle y\, =\, \dfrac{1}{3}x^3\, -\, 5x\, -\, \dfrac{4}{x}\)

...at which the tangent is horizontal.

---

I have not learned derivatives yet. I'm thinking I'm doing something with the fact that the slope of the tangent in this case is 0?

Thanks in advance!

That is correct. What did you get?

 

[amak]

That is correct. What did you get?

Ok great to hear I'm on the right track, but I'm stuck. I'm having a hard time simplifying this one.

 

[amak]

I'm stuck after that, so i set the difference quotient to zero but have trouble simplifying. Can anyone help?

 

[stapel]

Ok great to hear I'm on the right track, but I'm stuck. I'm having a hard time simplifying this one.

Please reply showing your work so far, so we can see what's going on. You set the right-hand side of the equation equal to zero, you multiplied through to clear all the fractions (noting that x cannot equal zero), noted that the resulting equation was quadratic in form, applied the Quadratic Formula, and... then what?

Please be complete. Thank you!