   SEARCH HOME Math Central Quandaries & Queries  Question from Cane, a student: Hi, my name is Cane and i am stuck on this problem. Find the domain and range then graph. y= square root of(x squared - 6x) + Square root of(6x - x squared) Thanks for the help Hi Cane,

I'm going to illustrate with the function

y = √(x2 - x) - √(x - x2)

For a number x to be in the domain of y you need both x2 - x ≥ 0 and x - x2 ≥ 0.

Think of the graph of y = x2 - x. If y = x2 - x changes from being above the x-axis (the function being positive) to being below the x-axis (the function being negative) the graph must pass through zero. Thus the function can only change sign at a point where x2 - x = 0, that is x = 0 or x = 1. This divides the x-axis into 3 segments, x < 0, 0 < x < 1, 1 < x . The expression (x2 - x) can't change sign in these segments since it can only change sign at x = 0 or x = 1. Now try a point in each segment.

At x = -1, x2 - x = 2 so x2 - x is positive for all x satisfying x < 0.
At x = 1/2, x2 - x = -1/2 so x2 - x is negative for all x satisfying 0 < x < 1.
At x = 2, x2 - x = 2 so x2 - x is positive for all x satisfying 1 < x.

Hence x2 - x ≥ 0 if x ≤ 0 or if x ≥ 1.

A similar argument with x - x2 shows that

x - x2 ≥ 0 if 0 ≤ x ≤ 1.

Thus to ensure both x2 - x ≥ 0 and x - x2 ≥ 0 x must satisfy x ≤ 0 or if x ≥ 1, and 0 ≤ x ≤ 1. The only values of x that qualify are x = 0 and x = 1. Thus the domain of y = √(x2 - x) - √(x - x2) consists of two points x = 0 and x = 1.

When x = 0, y = √(x2 - x) - √(x - x2) = 0 and when x = 1, y = √(x2 - x) - √(x - x2) = 0. Thus the range of y = √(x2 - x) - √(x - x2) consists of one point, y = 0.

I hope this helps,
Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.