Find the domain of the function using interval √x/(4x²+3x-1)
I solved 4x²+3x-1=0
I got (1/4,-1) but idk what to do
suppose we have y = √x/(4x²+3x-1)
There are two restrictions we have to worry about
1. The denominator can't be zero. You appear to have taken care of that
and found the values of x which make it zero
So we have to put on the restriction that x ≠ -1 and x ≠ 1/4
2. √x would be undefined if x is negative, so x ≥ 0
your domain would be all real values of x, x≥ 0 , x ≠ 1/4
(notice the x ≠ -1 is already taken care of with x ≥ 0)
Thanks
To find the domain of the function √x/(4x²+3x-1), we need to determine the x-values for which the function is defined. In this case, we need to consider two conditions:
1. The denominator should not be zero, as division by zero is undefined.
2. The value inside the square root (√x) should be non-negative (≥ 0), as the square root of a negative number is not a real number.
Let's address these conditions one by one:
1. To find when the denominator 4x²+3x-1 is zero, you correctly solved the quadratic equation 4x²+3x-1=0. You found the solution (1/4,-1). However, there is a mistake in the notation. The notation should be [1/4, -1]. There is no need to include round brackets since they imply open intervals. In this case, we need closed intervals to account for zero.
2. For the second condition, we need to find when √x ≥ 0. The square root of a number is non-negative (or zero) when the number itself is non-negative (or zero). Thus, x should be greater than or equal to zero, written as [0,∞).
Now, to find the intersection, or overlap, of the two conditions, we take the intersection of [1/4, -1] and [0,∞), which gives us the domain of the function:
Domain: [1/4, -1] ∩ [0,∞)
Simplifying further, we can see that the domain of the function is [1/4, ∞). Therefore, x-values between or equal to 1/4 and positive infinity are included in the domain of the function.