question 1 · 1 point\nconsider the graph of the function ( f(x)=\frac{x^{2}-1}{x^{2}+3 x - 4} ).\nfind the (…

question 1 · 1 point\nconsider the graph of the function ( f(x)=\frac{x^{2}-1}{x^{2}+3 x - 4} ).\nfind the ( x )-value of the removable discontinuity of the function.\nprovide your answer below:\nthe removable discontinuity occurs at ( x=) \ncontent attribution\nquestion 2 · 1 point

question 1 · 1 point\nconsider the graph of the function ( f(x)=\frac{x^{2}-1}{x^{2}+3 x - 4} ).\nfind the ( x )-value of the removable discontinuity of the function.\nprovide your answer below:\nthe removable discontinuity occurs at ( x=) \ncontent attribution\nquestion 2 · 1 point

Answer

Explanation:

Step1: Factor numerator and denominator

  • Numerator: (x^{2}-1=(x + 1)(x - 1)) (using (a^{2}-b^{2}=(a + b)(a - b)) with (a=x) and (b = 1)).
  • Denominator: (x^{2}+3x - 4=(x + 4)(x - 1)) (using (x^{2}+(a + b)x+ab=(x + a)(x + b)), where (a = 4) and (b=-1) since (4\times(-1)=-4) and (4+( - 1)=3)).

Step2: Simplify the function

The function (f(x)=\frac{(x + 1)(x - 1)}{(x + 4)(x - 1)}). For (x\neq1), we can cancel out the common factor ((x - 1)), and (f(x)=\frac{x + 1}{x + 4}) (the function is undefined at (x = 1) and (x=-4), but the discontinuity at (x = 1) is removable because the factor ((x - 1)) cancels).

Answer:

(1)