Question

question 1 · 1 point
consider the graph of the function ( f(x)=\frac{x^{2}-16}{x^{2}+5x - 36} ).
find the ( x )-value of the removable discontinuity of the function.
provide your answer below:
the removable discontinuity occurs at ( x=)

Explanation:

Step1: Factor numerator and denominator

For the numerator \(x^{2}-16=(x - 4)(x + 4)\) (using \(a^{2}-b^{2}=(a - b)(a + b)\) with \(a=x\), \(b = 4\)).
For the denominator \(x^{2}+5x-36=(x + 9)(x-4)\) (using \(x^{2}+bx + c=(x + m)(x + n)\) where \(m + n=b\) and \(mn=c\), here \(m = 9\), \(n=-4\)).
So \(f(x)=\frac{(x - 4)(x + 4)}{(x + 9)(x - 4)}\).

Step2: Simplify the function (for \(x

eq4\))
Cancel out the common factor \((x - 4)\) (since \(x
eq4\)), we get \(f(x)=\frac{x + 4}{x + 9}\) for \(x
eq4\).

Answer:

The removable discontinuity occurs at \(x = 4\)