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anonymous

  • 5 years ago

f(x) = x4 - 50x2 + 4 find the intervals on which f is increasing and decreasing. local min and max values. inflection points. intervals on which f is concave up and concave down.

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  1. amistre64
    • 5 years ago
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    1st and 2nd derivatives give you all that information

  2. amistre64
    • 5 years ago
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    y' = 4x^3 -100x y'' = 12x^2

  3. amistre64
    • 5 years ago
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    when y'=0 we get some critical points: y' = 4x^3 - 100x x(4x^2 - 100) = 0 x(2x-10)(2x+10) = 0 x = 0, x=5, x=-5 are all critical points to check <.......(-5)......(0).......(5).......>

  4. amistre64
    • 5 years ago
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    y'' = 12x^2 12(0)^2=0 ; good candidate for inflection 12(-5)^2 = 12(25) which is (+), this is a MIN 12(5)^2 = 12(25) which is (+), this is ALSO a MIN <.......(-5).......(0)........(5).......> ------0++++(0)++++0-----> This thing is a big "W" looking graph with -5, 0, 5 being the low, high, low respectively

  5. amistre64
    • 5 years ago
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    (-inf,0) concave up (0,inf) concave up (-5,5) concave down :) depends on where you lookat it from I suppose

  6. amistre64
    • 5 years ago
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    dont try to make any sense outta me line graph; it messed up anywhoos :)

  7. amistre64
    • 5 years ago
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    decreasing (-inf,-5) increaseing (-5,0) decreasing (0,5) increasing (5,inf)

  8. amistre64
    • 5 years ago
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    any of this making sense?

  9. amistre64
    • 5 years ago
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    f(-5) = (-5)^4 - 50(-5)^2 + 4 = (-5,-621) MIN f(0) = 0 - 0 + 4 = (0,4) MAX f(5) = (5)^4 - 50(5)^2 + 4 = (5,-621) MIN

  10. amistre64
    • 5 years ago
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    that should be all the answers :)

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