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anonymous

  • 5 years ago

Consider the function f(x) = x - x^3. determine one or more horizontal shifts that will change its form so that no linear term is present.

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  1. anonymous
    • 5 years ago
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    hmmmm

  2. anonymous
    • 5 years ago
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    Send x to x-c, collect terms and solve for c such that the coefficient of the linear term is exterminated.

  3. anonymous
    • 5 years ago
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    \[f(x-c)=(x-c)-(x-c)^3\]\[=(c^3-c)+(1-3c^2)x+3cx^2-x^3\]If you let \[c=\pm \frac{1}{\sqrt{3}}\]and sub. in, your linear (x) term will disappear.

  4. anonymous
    • 5 years ago
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    how did you know that?

  5. anonymous
    • 5 years ago
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    Translating a 1D function to the left or right amounts to shifting the argument by a constant. Just look at y=x^2 versus y=(x-1)^2.

  6. anonymous
    • 5 years ago
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    right, so you moved it so that the roots changed

  7. anonymous
    • 5 years ago
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    The question asks for a horizontal shift that will eliminate the linear term.

  8. anonymous
    • 5 years ago
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    ok , the original function has 3 roots. so shouldnt you move it up vertically ? oh wait... by translating it the roots will average out or something

  9. anonymous
    • 5 years ago
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    ok i see what you did, i got confused because it says shifts* plural , but that was clever

  10. anonymous
    • 5 years ago
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    It's just an application of the definition of 'horizontal shift' along with the condition stipulated in the question that the shift is wanted so that the linear term is extinguished.

  11. anonymous
    • 5 years ago
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    It's a bizarre question.

  12. anonymous
    • 5 years ago
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    you know anything about gabriel horn

  13. anonymous
    • 5 years ago
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    the paradox is that there is no bottom to the horn, so in theory you can keep adding say water to it (ideal water) . but the volume is finite

  14. anonymous
    • 5 years ago
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    ideal water (quantum molecular issues aside)

  15. anonymous
    • 5 years ago
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    It is indeed a weird question, but a similar technique can be used to reduce, for example, a quartic equation to one with no cubic term which makes it easier to solve in some ways. Sounds little random but that's just about application of this thing I've seen before.

  16. anonymous
    • 5 years ago
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    Yes

  17. anonymous
    • 5 years ago
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    Same with suppressing a cubic.

  18. anonymous
    • 5 years ago
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    here is another crazy integral. if you find the surface area of y= ln x from 0 to 1 (so its below x axis), and revolve it about y axis , you get a finite surface area (even though y = ln x goes to negative infinity)

  19. anonymous
    • 5 years ago
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    its called depressed equation , but what you did is straightforward :)

  20. anonymous
    • 5 years ago
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    actually there might be another way, buts kind of long. you can use the symmetric polynomial reduction form . so x^3 - (a+b+c) x^2 + (ab +ac +bc)x - abc , where a,b,c are your roots

  21. anonymous
    • 5 years ago
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    yes, depressed.

  22. anonymous
    • 5 years ago
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    so you want ab + ac + bc = 0

  23. anonymous
    • 5 years ago
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    nevermind, you have it :)

  24. anonymous
    • 5 years ago
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    On the bright side, I like discussions on here with people who can do Maths, and don't spam exactly the same question over and over and do nothing themselves - makes a nice change.

  25. anonymous
    • 5 years ago
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    yup

  26. anonymous
    • 5 years ago
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    newton can you look over my solution for work problem, im stuck on the density g thing

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