Consider the function f(x)=x
2+c
Let P=(x,y) be a point on the graph of y=x
2+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
Point P can be expressed as (x,x^2+c)
Then D(x) = sqrt[x^2 + (x^2+c)^2]
or
D(x) = sqrt[x^4 + (2c+1)x^2 + c^2]
"[2] Prove D(x) >= F(x) for all x and c."
sqrt[x^2 + (x^2+c)^2] >= x^2+c
x^2 + (x^2+c)^2 >= (x^2+c)^2
x^2 >= 0
This is trivially true, so [2] is proven.
"[1] Find the value(s) of x, in terms of c, that minimizes D(x)."
D'(x) = (4x^3 + (4c+2)x)/(2*sqrt[x^4 + (2c+1)x^2 + c^2])
Only the numerator needs to be considered for finding zeros:
4x^3 + (4c+2)x = 0
x*(4x^2 + 4c+2) = 0
x = 0 or x = sqrt[-c - 1/2]
For c >= 1/2 there is only one option: D(0) = abs(c)
For c < -1/2 the minimul could be D(0) = abs(c) or D(sqrt[-c - 1/2]) = sqrt[-c-1/2 + (-c-1/2 + c)^2] = sqrt[-c - 1/4]
In this case sqrt[-c - 1/4] is never larger:
sqrt[-c - 1/4] < abs(c)
-c - 1/4 =< c^2
0 =< c^2 + c + 1/4
0 <= (c + 1/2)^2
Then the answer to [1] is: x=0 for c>=-1/2 and x=sqrt[-c-1/2] for c<-1/2
"[3] Find, in terms of c, the limit as x goes to infinity of d(x)-f(x)"
Lim {x -> +inf} D(x)-F(x)
= Lim {x -> +inf} sqrt[x^4 + (2c+1)x^2 + c^2] - (x^2+c)
This is indeterminate of the form infinity - infinity. Put the entire expression over 1 and multiply the numerator and denominator by x^-2:
= Lim {x -> +inf} (sqrt[1 + (2c+1)/x^2 + c^2/x^4] - 1 - c/x^2) / (x^-2)
Now the limit is of the form 0/0 and L'Hopital's Rule can be applied:
= Lim {x -> +inf} [(-2(2c+1)/x^3 + -4c^2/x^5)/(2*sqrt[1 + (2c+1)/x^2 + c^2/x^4]) + 2c/x^3]/(-2x^-3)
Multiply numerator and denominator by x^3:
= Lim {x -> +inf} [(-2c-1-2c^2/x^2)/sqrt[1 + (2c+1)/x^2 + c^2/x^4] + 2c]/(-2)
= [(-2c-1-0)/sqrt[1+0+0] + 2c]/(-2)
= 1/2
For this problem the limit in [3] is always 1/2
Parts 1-3
