Let's analyze each function to determine if the discontinuity at the specified $a$ is removable or not. A discontinuity is removable if we can redefine the function at $a$ such that the function becomes continuous at that point.

---

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

We need to check if there is a removable discontinuity at $x = 1$.

• Step 1: Factor the numerator.
  $x^4 - 1$ is a difference of squares: $x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$

• Step 2: Simplify the function.
  Now, rewrite $f(x)$: $f(x) = \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1}$ For $x \neq 1$, we can cancel out $(x - 1)$: $f(x) = (x + 1)(x^2 + 1)$

• Step 3: Define $g(x)$, which agrees with $f(x)$ for $x \neq 1$ and is continuous at $x = 1$.
  Since the function simplifies to $g(x) = (x + 1)(x^2 + 1)$, we can evaluate $g(1)$: $g(1) = (1 + 1)(1^2 + 1) = 2 \times 2 = 4$ Therefore, the discontinuity is removable, and the function $g(x)$ is: $g(x) = (x + 1)(x^2 + 1)$

---

(b) $f(x) = \frac{x^3 - x^2 - 12x}{x - 4}$, $a = 4$

We need to check if there is a removable discontinuity at $x = 4$.

• Step 1: Factor the numerator.
  $x^3 - x^2 - 12x$ can be factored by first factoring out $x$: $x^3 - x^2 - 12x = x(x^2 - x - 12)$ Now factor $x^2 - x - 12$: $x^2 - x - 12 = (x - 4)(x + 3)$ So, we have: $f(x) = \frac{x(x - 4)(x + 3)}{x - 4}$

• Step 2: Simplify the function.
  For $x \neq 4$, we can cancel out $(x - 4)$: $f(x) = x(x + 3)$

• Step 3: Define $g(x)$, which agrees with $f(x)$ for $x \neq 4$ and is continuous at $x = 4$.
  Since the function simplifies to $g(x) = x(x + 3)$, we can evaluate $g(4)$: $g(4) = 4(4 + 3) = 4 \times 7 = 28$ Therefore, the discontinuity is removable, and the function $g(x)$ is: $g(x) = x(x + 3)$

---

(c) $f(x) = \sin(x)$, $a = \pi$

• Step 1: Analyze the function.
  The function $f(x) = \sin(x)$ is continuous everywhere in its domain, including at $x = \pi$. There is no discontinuity at $x = \pi$.

Thus, the discontinuity is not removable, and the function $g(x)$ does not need to be redefined since there is no discontinuity.

---

Final Answers:

• (a) The discontinuity is removable.
  $g(x) = (x + 1)(x^2 + 1)$

• (b) The discontinuity is removable.
  $g(x) = x(x + 3)$

• (c) The discontinuity is not removable (since there is no discontinuity).
  $g(x) = \text{DNE}$

---

Do you have any questions or would you like further details?

Here are 5 related questions to deepen understanding:

1. How do you factor a cubic polynomial to find removable discontinuities?
2. What is the difference between removable and non-removable discontinuities?
3. How do you check if a function is continuous at a point?
4. Can all rational functions have removable discontinuities?
5. What role do limits play in determining removable discontinuities?

Tip: Always check if you can simplify a function by factoring when looking for removable discontinuities.