question 1 · 1 point\nconsider the graph of the function ( f(x)=\frac{x^{2}-16}{x^{2}+5x - 36} ).\nfind the…

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

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

Answer

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\neq4))

Cancel out the common factor ((x - 4)) (since (x\neq4)), we get (f(x)=\frac{x + 4}{x + 9}) for (x\neq4).

Answer:

The removable discontinuity occurs at (x = 4)