Determine the values of x for which the function f(x) = 3/(√x^2+3x-4) is continuous.
All values of x except where the denominator vanishes, which happens at -4 and 1
Can
You please explain the steps
To determine the values of x for which the function f(x) = 3/(√x^2+3x-4) is continuous, we need to consider two conditions:
1. The denominator (√x^2+3x-4) should not be zero.
2. The function should not have any points of discontinuity, such as vertical asymptotes or removable discontinuities.
Let's begin with the first condition:
1. The denominator (√x^2+3x-4) should not be zero.
To find when the denominator is zero, we set (√x^2+3x-4) equal to zero and solve for x.
√x^2+3x-4 = 0
By squaring both sides to eliminate the square root:
x^2 + 3x - 4 = 0
Now we can factor the quadratic equation:
(x + 4)(x - 1) = 0
Setting each factor to zero, we find two potential values for x:
x + 4 = 0 or x - 1 = 0
From these equations, we get:
x = -4 or x = 1
So, the denominator is zero at x = -4 and x = 1.
Now let's consider the second condition:
2. The function should not have any points of discontinuity.
In this case, we need to check if there are any vertical asymptotes or removable discontinuities in the function.
The function f(x) = 3/(√x^2+3x-4) has a vertical asymptote when the denominator approaches zero but the numerator does not. We already found that the denominator is zero at x = -4 and x = 1. To check if there are vertical asymptotes, we need to consider the behavior of the numerator.
In this case, since the constant 3 is divided by the denominator, the numerator does not approach infinity or negative infinity as x approaches -4 or 1. Therefore, there are no vertical asymptotes.
Additionally, the function f(x) does not have any removable discontinuities since the numerator and the denominator are both continuous functions.
In conclusion, the function f(x) = 3/(√x^2+3x-4) is continuous for all values of x except x = -4 and x = 1.