Find a such that min {x2,4} = x2 x 1[0,a] (x) + 4 x 1(a,8) (x)

Question

Find a such that min {x²,4} = x² x 1_[0,a] (x)  +  4 x 1_(a,∞) (x)

 

 

1[0,a] (x) basically means that if X is within the intervals that the answer will be 1 and if it is not then the answer will be 0.

However I cannot seem to be able to work out a.

 

Answer 1

Let's examine the expression on the right of the equals sign.  We see that [0,a] and (a, infinity) are mutually exclusive, and together form a complete interval [0, infinity).  This allows us to treat the expression as a composite function.

 

When x<0, 1[0,a](x) = 0 and 1(a, infinity)(x) = 0.  So the expression is 0.

When 0 <= x <= a, 1[0,a] = 1 and 1(a, infinity)(x) = 0.  So the expression is x^2.

When x > a, 1[0,a](x) = 0 and 1(a, infinity)(x) = 1.  So the expression is 4.

Also, notice that min {x^2, 4} cannot be zero when x < 0, so it seems that we need to restrict our overall domain of x to [0, infinity). This allows us to neglect the first interval when describing the composite function (x cannot be negative).

 

Let's call this composite function g(x), and write it as

 

g(x) = x^2, 0 <= x <= a. (x is greater than or equal to 0, and less than or equal to a).

g(x) = 4, x > a.

 

Rewriting the equation for the first interval, we have

min{x^2,4} = x^2, x <= a. 

min {x^2,4} = x^2 means that x^2 is smaller than 4.  This is true for x <= 2, so a = 2.

 

Rewriting the equation for the second interval, we have

min{x^2,4} = 4, x > a.  min{x^2,4} = 4 means that x^2 >= 4.  This is true for x > 2, so a = 2.

 

Since a = 2 satisfies the requirements on both intervals [0,a] and (a, infinity), a = 2 is the answer.

 

Let's check by substituting a = 2 into g(x) and see if the results are valid:

min{x^2,4} = x^2, 0 <= x <= 2.  This checks, since x^2 <= 4 if x <= 2.

min{x^2,4} = 4, x>2.  This also checks, since 4 < x^2 if x >2.