Find the interval (or intervals) on which the given expression is defined:
(x^2-7x+12)^(1/2)
we can write it as
√[(x-3)(x-4) ]
remember that we cannot take √ of a negative number
the critical values above are x=3 and x=4
for values between 3 and 4, the result is negative,
e.g. x = 3.5 , we get √(.5(-.5)) = √-.25 which is undefined.
so we need
x ≤ 3 OR x ≥ 4
Wouldn't they both be > or = to?
no, take an example of x ≤ 3
e.g. x = -5
expression is
√(-8)(-9) = √72 which is a real number
Trust me, my answer is correct.
Yes you're right. THANK YOU SO MUCH!!
To find the interval(s) on which the given expression is defined, we need to consider the domain of the square root function.
The square root function is defined for non-negative values only, since taking the square root of a negative number would result in a complex number.
In the given expression, we have (x^2 - 7x + 12)^(1/2). To find the interval(s), we need to determine the values of x for which the expression inside the parentheses is non-negative.
Let's solve the expression inside the parentheses for x:
x^2 - 7x + 12 = 0
We can factor this quadratic equation:
(x - 3)(x - 4) = 0
Setting each factor equal to zero, we find that x = 3 or x = 4.
To determine the interval(s) on which the expression is defined, we consider the ranges of values for x.
Since we have (x^2 - 7x + 12)^(1/2) in the expression, we need to find the values of x for which (x^2 - 7x + 12) is non-negative.
Analyzing the quadratic equation, we see that it is a parabola opening upwards. The roots of the equation are x = 3 and x = 4.
Considering the interval between the roots, we find that the expression is defined for x values between 3 and 4, inclusive (i.e. x ∈ [3,4]).
Therefore, the interval on which the given expression is defined is [3, 4].