Calculus

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if f(x)= |(x^2-9)(x^2+1)| how many numbers in the interval [-1,1] satisfy the mean value theorem?
None
1
2
3
Will someone please explain in detail?

  • Calculus -

    absolute value problems can get tricky, because the functions are not differentiable at any cusps.

    However, since (x^2-9)(x^2+1) is negative on [-1,1],

    f(x) = -(x^2-9)(x^2+1)

    This is a nice smooth polynomial, so it satisfies the requirements of the MVT on [-1,1].

    Now take a look at the graph:

    http://www.wolframalpha.com/input/?i=%7C(x%5E2-9)(x%5E2%2B1)%7C

    Since f(-1) = f(1) we see that there is a single point where f'(c) = 0 on [-1,1].

    To show that algebraically, we just go through the steps:

    (f(1)-f(-1))/2 = 0
    f'(x) = 4x(4-x^2)
    f'(0) = 0

  • Calculus -

    so the answer is none?

  • Calculus -

    Excuse me?
    Did you look at the graph?
    Did you actually read what I wrote?

    I don't mind helping you arrive at a solution, but I do expect you to actually read it, rather than just hold out your hand and say "gimme the answer." In fact, I did give you the answer.

  • Calculus -

    Yes I read it I still don't get it and 0 isn't an answer choice.

  • Calculus -

    That is because x=0 is the single point where f'(x) = 0

    the number zero is a single choice, not an absence of choices!

    The answer is 1

    There is 1 (one) point on the interval [-1,1] where

    f(1)-f(-1)
    ------------ = 0
    1 - (-1)

    That point is at x=0.

    Just take a deep breath and read what is written.

  • Calculus -

    wait its 3 right?

  • Calculus -

    oh ok thank you!!!! sorry

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