If f(x)= sqrt(9x^2-12x+4)/(15x^2+2x-8),what is d domain of the function. And what is it in absolute value(abs)
To find the domain of a function, we need to identify any values of x that would make the function undefined. In this case, we have a square root in the numerator and a denominator that could potentially equal zero at certain values of x.
The domain of the function is the set of all real numbers that satisfy the following conditions:
1. The expression inside the square root (the radicand) must be greater than or equal to zero to avoid taking the square root of a negative number.
2. The denominator should not be equal to zero as it would result in division by zero.
Let's solve these conditions step by step:
Condition 1: The radicand (9x^2 - 12x + 4) must be greater than or equal to zero.
In order to determine the values of x that satisfy this condition, we can find the roots of the quadratic equation 9x^2 - 12x + 4 = 0.
To do this, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 9, b = -12, and c = 4. Plugging these values into the formula, we get:
x = (-(-12) ± √((-12)^2 - 4(9)(4)))/(2(9))
= (12 ± √(144 - 144))/(18)
= (12 ± √0)/(18)
= 12/18 = 2/3
Hence, the quadratic equation has a single root, x = 2/3. This means that for any value of x less than or greater than 2/3, the radicand will be positive. Therefore, the radicand is greater than or equal to zero for all x in the domain.
Condition 2: The denominator (15x^2 + 2x - 8) should not equal zero.
To find the values of x for which the denominator is equal to zero, we can solve the quadratic equation 15x^2 + 2x - 8 = 0.
Using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac))/(2a)
The coefficients in this case are: a = 15, b = 2, and c = -8. Plugging these values into the formula gives:
x = (-(2) ± √((2)^2 - 4(15)(-8)))/(2(15))
= (-2 ± √(4 + 480))/(30)
= (-2 ± √484)/(30)
= (-2 ± 22)/(30)
Hence, the solutions are x = 1/3 and x = -8/5.
Now, we can determine the domain of the function:
1. The radicand (9x^2 - 12x + 4) is always greater than or equal to zero for all real values of x.
2. The denominator (15x^2 + 2x - 8) is not equal to zero for all real values of x except x = 1/3 and x = -8/5.
Therefore, the domain of the function f(x) = sqrt(9x^2 - 12x + 4)/(15x^2 + 2x - 8) is all real numbers except x = 1/3 and x = -8/5.
To express the domain in absolute value form, we can write:
|x| ≠ |1/3|, |x| ≠ |8/5|
In other words, the absolute value of x should not be equal to the absolute value of 1/3 or the absolute value of 8/5.