If f(x)=x+2/x^2-9 and g(x)=11/x^2+3x
(a) find f(x)+g(x)
(b) list all of the excluded values
(c) classify each type of discontinuity
What is being asked in the problem, what plan and I going to try?
Solve step by step
(a) To find f(x)+g(x), we first need to add the two functions:
f(x) + g(x) = (x+2)/(x^2-9) + 11/(x^2+3x)
Next, find a common denominator for the two fractions:
f(x) + g(x) = (x+2)(x^2+3x) / ((x^2-9)(x^2+3x)) + 11(x^2-9) / ((x^2-9)(x^2+3x))
Now, combine the fractions:
f(x) + g(x) = (x^3 + 3x^2 + 2x + 11x^2 - 99) / (x^4 + 3x^3 - 9x^2 - 27x)
Simplify the numerator:
f(x) + g(x) = (x^3 + 14x^2 + 2x - 99) / (x^4 + 3x^3 - 9x^2 - 27x)
Therefore, f(x) + g(x) = (x^3 + 14x^2 + 2x - 99) / (x^4 + 3x^3 - 9x^2 - 27x)
(b) To find the excluded values, we need to identify where the denominator of the functions equals zero. In this case, we have two denominators: x^2 - 9 = 0 and x^2 + 3x = 0.
Solving x^2 - 9 = 0, we get x = ±3 as excluded values.
Solving x^2 + 3x = 0, we get x(x+3) = 0, so x = 0 and x = -3 are the excluded values.
Therefore, the excluded values are x = ±3, 0.
(c) To classify the types of discontinuity, we need to look at the behavior of the functions at the excluded values.
At x = 3 and x = -3, the functions have vertical asymptotes because the denominators approach zero.
At x = 0, the functions exhibit a removable discontinuity because the factor (x) cancels out from the numerator and denominator.