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I can see the origin from here. (Posted on 2012-03-11) Difficulty: 4 of 5
Consider the function f(x)=x2+c
Let P=(x,y) be a point on the graph of y=x2+c
Let d(x) = the distance from P to the origin.

[1] Find the value(s) of x, in terms of c, that minimizes d(x).

[2] Prove d(x) ≥ f(x) for all x and c.

[3] Find, in terms of c, the limit as x goes to infinity of d(x)-f(x)

[4] Find, in terms of c, and if it exists, the integral from negative infinity to positive infinity of (d(x)-f(x)-the above limit) dx

  Submitted by Jer    
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Solution: (Hide)
[1]
If c ≥ -1/4 then x=0
If c < -1/4 then x = √((2c+1)/-2)

[2]
Connect P to O. The hypotenuse is always the longest.

[3]
The answer is 1/2 for any c.
We want √(x2+(x2+c)2) - (x2+c)
Multiply by the conjugate and its reciprocal: (x2+(x2+c)2-(x2+c)2) / √(x2+(x2+c)2) + (x2+c)
x2 / [x2((1/x2)√(x2+(x2+c)2) + (1/x2)(x2+c))]
1 / [√(1/x2 + 1 + 2c/x2 + c2/x2) + 1 + c/x2]
The limit as x goes to infinity is 1 / [√(0+1+0+0) + 1+0] = 1/2

[4]
Don't know yet, but if c = -4 the answer seems to be about 23.75 I dare you to put sqrt(x^2+(x^2+c)^2) - (x+c)^2 - 1/2 into the wolfram integrator. When c=0 it's not hard to show the answer is -2/3.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Part 4 still unsolvedJer2012-10-05 10:57:06
Some ThoughtsParts 1-3Brian Smith2012-03-13 20:18:30
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