SOLUTION: Given that a,b, and c are constants, a<b<c, state the domain
y= sqrt((x-a)(x-b)(x-c))
Hint: Graph y=(x-a)(x-b)(x-c)
Algebra.Com
Question 1133768: Given that a,b, and c are constants, a
y= sqrt((x-a)(x-b)(x-c))
Hint: Graph y=(x-a)(x-b)(x-c)
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
For the square root of an expression to be a real number, the expression has to be zero or positive.
In this example, the expression under the radical is (x-a)(x-b)(x-c), and we are given the condition that a < b < c.
That expression has the value 0 when x is a, b, or c; so the values a, b, and c are all in the domain.
For the intervals into which the rest of the number line is divided by those three values...
if x > c, then all three factors in the expression are positive, so the product is positive;
if b < x < c, then the factor (x-c) is negative, so the product is negative;
if a < x < b, then the factors (x-b) and (x-c) are both negative, so the product is positive; and
if x < a, then all three factors are negative, so the product is negative.
Now you should be able to determine the domain....
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