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Schrodinger Group Title

Followup to a derivatives question earlier: I'm told to find all critical points and specify which extrema are maxima or minima. I'm having trouble just chunking through the numbers. (Problem below).

  • one year ago
  • one year ago

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  1. stamp Group Title
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    i am not a doctor but i will see what i can help you

    • one year ago
  2. Schrodinger Group Title
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    \[y = x ^{2/3}(x+2)\]\[y \prime = \frac{ 5x+4 }{ 3\sqrt[3]{x} }\]All critical points should occur at an endpoint or at y' = 0 or y' being undefined by the definition of a critical point. The answers that satisfy the definition are \[x = 0, x = -\frac{ 4 }{ 5 }.\]Plugging the x values back into y itself, I am supposed to get\[y = 0, y =\frac{ 12 }{ 25 }10^{1/3}\]I have no idea how that second one is gotten. First one, makes sense, everything is cancelled out by being multiplied by zero. Second one, you have \[(-\frac{ 4 }{ 5 })^{2/3}(-\frac{ 4 }{ 5 }+2)\]How the hell are they ending up with that neat, fractional answer with that last bit?

    • one year ago
  3. Schrodinger Group Title
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    (Oh, and pardon me, i'm just horrible at algebra. Always have, will be marginally better with time. Eventually. Maybe.)

    • one year ago
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    y'(0) does not exist 5x + 4 = 0 when x = -4/5 so y'(-4/5) = 0 y'(-4/5) is your critical point

    • one year ago
  5. Schrodinger Group Title
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    y'(0) is still a critical point. It's undefined, but it doesn't not exist. And by the definition of a critical point [y'(0) is undefined, an end point within a defined interval or at a point where y' = 0], it is a critical point. My book lists it as an answer word for word, and my book is definitely right.

    • one year ago
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    Well if x = 0 and x = -4/5 are you critical points, find f(0) and f(-4/5)

    • one year ago
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    f(0) = 0

    • one year ago
  8. Schrodinger Group Title
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    I found them above. I'm just having trouble numerically working out the second one.

    • one year ago
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    something is amiss

    • one year ago
  10. Schrodinger Group Title
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    What's amiss? I'm just bad with fractions being raised to fractional exponents less than one, especially when the base is also less than one, lol. I just don't understand how they're getting a clean, fractional answer from it. Whenever I take the cube root of those fractions I just get an irrational slew of numbers, not a clean-cut, dandy little fraction.

    • one year ago
  11. phi Group Title
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    \[ (-\frac{ 4 }{ 5 })^{2/3}(-\frac{ 4 }{ 5 }+2) \] \[ (\frac{ 4^2 }{ 5^2 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ (\frac{ 2\cdot 2^3 }{ 5^2 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ (\frac{ 5\cdot 2\cdot 2^3 }{ 5 \cdot 5^2 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ (\frac{ 10\cdot 2^3 }{ 5^3 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ 10^{1/3}(\frac{ 2 }{ 5 })(\frac{ 6 }{ 5 }) \] \[ 10^{1/3}\frac{ 12 }{ 25 }\]

    • one year ago
  12. Schrodinger Group Title
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    @phi , Line 4, did you just multiply the left portion by 5/5?

    • one year ago
  13. phi Group Title
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    yes, because it would be nice to have 5^3 in the bottom, so we can take its cube root.

    • one year ago
  14. Schrodinger Group Title
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    Makes sense, just making sure. Thanks so much, dude, this helps a lot.

    • one year ago
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