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Anonymous
Anonymous asked in Science & MathematicsMathematics · 10 years ago

help!calculus!concavity!?

Test f(x)=2x^4-5x^3+5 for concavity and inflection points.

a) f has inflection points when x=____and x=_____(list the smaller value first)

b) f is concave up on ( -∞,__) ∪ (___,+∞) and its concave down on (__ , __)

thx all!!!

4 Answers

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  • JJ
    Lv 7
    10 years ago
    Favorite Answer

    Not sure exactly, but concave is like this: (__)

    and convex is like this: )__(

    Hope that helps.

  • 10 years ago

    f(x) = 2x⁴ - 5x³ + 5

    f'(x) = 8x³ - 15x²

    f"(x) = 24x² - 30x

    (using 2nd derivative to determine concavity and inflection points)

    set the second derivative to zero and solve for x, the inflection points

    24x² - 30x = 0

    6x(4x-5) = 0

    6x = 0

    x = 0

    4x-5 = 0

    4x = 5

    x = 5/4

    picking any point in between -∞ and 0, 0 and 5/4, 5/4 and ∞ to test concavity.

    if it's negative it's concave down and if it's costive it's concave up.

    in this case

    -∞ < -1 < 0 → f"(-1) > 0 so (-∞,0) is concave up

    0 < 1 <5/4 → f"(1) < 0 so (0,5/4) is concave down

    5/4 < 2 < ∞ → f"(2) > 0 so (5/4,∞) is concave up

    (-∞,0) ∪ (5/4,∞) concave up

    (0,5/4) is concave down

    inflection points at x = 0 and x = 5/4

  • 10 years ago

    Find the double prime of the function, which is f ''(x) = 24x^2-30x. Set that to zero to get your inflection points and concavity.

    a) (0,5) and (5/4, 15/128)

    b) up on (-∞,0) ∪(5/4,+∞) and down on (0, 5/4)

    Source(s): Accounting major
  • Anonymous
    10 years ago

    http://en.wikipedia.org/wiki/Inflection_point

    If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists, but this condition does not provide a sufficient definition of a point of inflection

    Read that, have a go. Take values either side to check if its going up or down.

    Lastly, plot a few values to see what the function looks like on paper. Good luck.

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