A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 4 years ago

find all points on the curve y=x^3-3X where the tangent is parallel to the x-axis.

  • This Question is Open
  1. calculusfunctions
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    When in doubt always start with the derivative of the function. Hence if y = x³ − 3x dy/dx = 3x² − 3 where dy/dx is of course the instantaneous rate of change function (the slope function). Since the tangent to y is parallel to the x-axis, dy/dx = 0 ∵ the slope of the x-axis is zero (x-axis is horizontal and the slope of any horizontal line is zero). Thus dy/dx = 0 and since dy/dx = 3x² − 3, 3x² − 3 = 0 solving for x, we obtain x = ±1 when x = -1, y = 2 when x = 1, y = -2 ∴ the tangent to y = 3x² − 3 is parallel to the x-axis at the points (-1, 2) and (1, -2)

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.