Question

question 1 · 1 point
consider the graph of the function ( f(x)=\frac{x^{2}-1}{x^{2}+3 x - 4} ).
find the ( x )-value of the removable discontinuity of the function.
provide your answer below:
the removable discontinuity occurs at ( x=)
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question 2 · 1 point

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 \(x
eq1\), 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:

\(1\)