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