Could anyone please help me figure out the discontinuity that is removable of the following function?? Any help would be greatly appreciated! Thanks in advance!!!
f(x)=4x/x^2+x-2
x^2+x-2=0
x=1 is your discontinuity
Why would it be 1?? I'm so confused...
because having 0 in the denominator would make your function be undefined and therefore be a discontinuity at that point. If you set the denominator equal to 0 and solve your answer would be 1.
x^2+x-2=0
(1)^2 + 1 - 2=0
1+1-2=0
To determine if there is a removable discontinuity in the function f(x) = 4x/(x^2 + x - 2), we need to check if there is a value of x that makes the denominator equal to 0. If we find such a value, it means there might be a removable discontinuity at that point.
In this case, the denominator of the fraction is x^2 + x - 2. To find the values of x that make the denominator equal to 0, we can set the denominator equal to 0 and solve for x:
x^2 + x - 2 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's factor it:
(x + 2)(x - 1) = 0
Now we have two values of x that make the denominator equal to 0: x = -2 and x = 1.
So, the potential removable discontinuities occur at x = -2 and x = 1. To determine if these are removable discontinuities, we need to check if the limit of f(x) exists as x approaches -2 and x approaches 1.
To find the limits, substitute the x-values into the function and simplify:
When x approaches -2:
f(-2) = 4(-2)/((-2)^2 + (-2) - 2) = -8/0
When x approaches 1:
f(1) = 4(1)/((1)^2 + 1 - 2) = 4/0
Since both limits result in a division by 0, we have an indeterminate form. This means that there is a potential removable discontinuity at both x = -2 and x = 1. However, to confirm if these are actually removable discontinuities, further analysis and investigation, such as graphing or evaluating the function at those points, may be needed.