Question Statement

Find the points of discontinuity for the function:

$f(x)=\frac{x^2-1}{x^4-1}$

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Background and Explanation

A function is continuous at a point only if it is defined there. For rational functions (fractions with polynomials), discontinuities occur where the denominator equals zero, making the function undefined.

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Solution

To determine where $f(x)$ is discontinuous, we identify where the function is undefined by finding when the denominator equals zero.

Step 1: Find where the denominator is zero

Set the denominator equal to zero: $x^4-1=0$

Solving for $x$: $x^4=1$ $x=\pm 1$

Step 2: Verify the function is undefined

At $x=1$:

• Numerator: $1^2-1=0$
• Denominator: $1^4-1=0$

At $x=-1$:

• Numerator: $(-1{)}^2-1=0$
• Denominator: $(-1{)}^4-1=0$

Since the denominator equals zero at both points, $f(x)$ is not defined at $x=\pm 1$.

Step 3: Identify points of discontinuity

Because $f(x)$ is undefined at $x=1$ and $x=-1$, the function is discontinuous at these points.

Therefore, the points of discontinuity are: $x=-1\text{and}x=1$

(Note: These are removable discontinuities. By factoring $f(x)=\frac{(x-1)(x+1)}{(x-1)(x+1)(x^2+1)}$, we see the function simplifies to $\frac{1}{x^2+1}$ for $x\ne \pm 1$, indicating holes in the graph at these points rather than vertical asymptotes.)

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Key Formulas or Methods Used

• Condition for discontinuity: A rational function is discontinuous where its denominator equals zero (making it undefined)
• Difference of squares: $a^2-b^2=(a-b)(a+b)$, used to factor $x^2-1$ and $x^4-1$
• Domain restriction: $x^4-1=0$ yields $x=\pm 1$

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Summary of Steps

1. Set the denominator equal to zero: $x^4-1=0$
2. Solve for $x$ to find where the function is undefined: $x=\pm 1$
3. Conclude that $f(x)$ is discontinuous at $x=-1$ and $x=1$