anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
hmmmm
anonymous
  • anonymous
Send x to x-c, collect terms and solve for c such that the coefficient of the linear term is exterminated.
anonymous
  • anonymous
\[f(x-c)=(x-c)-(x-c)^3\]\[=(c^3-c)+(1-3c^2)x+3cx^2-x^3\]If you let \[c=\pm \frac{1}{\sqrt{3}}\]and sub. in, your linear (x) term will disappear.

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anonymous
  • anonymous
how did you know that?
anonymous
  • anonymous
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.
anonymous
  • anonymous
right, so you moved it so that the roots changed
anonymous
  • anonymous
The question asks for a horizontal shift that will eliminate the linear term.
anonymous
  • anonymous
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
anonymous
  • anonymous
ok i see what you did, i got confused because it says shifts* plural , but that was clever
anonymous
  • anonymous
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.
anonymous
  • anonymous
It's a bizarre question.
anonymous
  • anonymous
you know anything about gabriel horn
anonymous
  • anonymous
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
anonymous
  • anonymous
ideal water (quantum molecular issues aside)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Same with suppressing a cubic.
anonymous
  • anonymous
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)
anonymous
  • anonymous
its called depressed equation , but what you did is straightforward :)
anonymous
  • anonymous
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
anonymous
  • anonymous
yes, depressed.
anonymous
  • anonymous
so you want ab + ac + bc = 0
anonymous
  • anonymous
nevermind, you have it :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
yup
anonymous
  • anonymous
newton can you look over my solution for work problem, im stuck on the density g thing

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