how to find turning points of a quartic function using calculus?

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There are at most three turning points for a quartic, and always at least one.

At a turning point (of a differentiable function) the derivative is zero. However the derivative can be zero without there being a turning point. (Consider $f(x)=x^3$ or $f(x)=x^5$ at $x=0$).

A good strategy for kinds of functions you don't completely understand is to sketch them - this works well for polynomials. What is the general shape of a quartic with positive coefficient of $x^4$?

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Nyx Smith
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Nyx Smith

Updated on August 01, 2022

Comments

  • Nyx Smith
    Nyx Smith 3 months

    My teacher gave a function of $$f(x)= 2x^4-3x^3-21x^2+16x+60$$ He said that there needs to be $4$ turning points. I only now how to find the turning points if the function is at cubic not quartic.

    • A. Goodier
      A. Goodier about 3 years
      Did you try differentiating and setting the derivative $=0$?
    • Nyx Smith
      Nyx Smith about 3 years
      I'm not sure. This topic is new so I really have no idea what to do
    • A. Goodier
      A. Goodier about 3 years
      Do you know how to work out $f'(x)$?
    • Nyx Smith
      Nyx Smith about 3 years
      No. I'm pretty hopeless at this. I'm not really good with math or other terms related to it.
    • A. Goodier
      A. Goodier about 3 years
      To be honest, it is difficult to help you with this if you do not know how to differentiate. I suggest you practice differentiating simple functions on Khan Academy or similar before attempting this question.
    • Toby Mak
      Toby Mak about 3 years
      What is your definition of 'turning point'? Many people have interpreted it as a point of 'inflection point' where $f''(x) = 0$, but you can also interpret it as where the function 'turns' or where $f'(x) = 0$.
  • Nyx Smith
    Nyx Smith about 3 years
    what method do i use in order to get 3 x-coordinates?
  • Toby Mak
    Toby Mak about 3 years
    You have that $x=2$ by the rational root theorem. Dividing $f'(x)$ by $x-2$ gives the other two roots, since the other factor is a quadratic. You should use the quadratic formula to do this, since you will find the quadratic cannot be factored.
  • Angina Seng
    Angina Seng about 3 years
    I get $f'(x)=8x^3-9x^2-42x+16$.
  • Toby Mak
    Toby Mak about 3 years
    That was a silly mistake, thanks. The rest of the post should not be affected.
  • Angina Seng
    Angina Seng about 3 years
    I don't get $f'(2)=0$.