Question

(a) ( f(x) = \frac{x^4 - 1}{x - 1}, quad a = 1 )
( \bigcirc ) the discontinuity is removable.
( \bigcirc ) the discontinuity is not removable.
( g(x) = square )
(b) ( f(x) = x^3 - x^2 - 56x )

Explanation:

Step1: Factor the numerator

We know that \(x^4 - 1\) is a difference of squares, which can be factored as \((x^2 + 1)(x^2 - 1)\). And \(x^2 - 1\) is also a difference of squares, so we can further factor it as \((x^2 + 1)(x + 1)(x - 1)\). So the function \(f(x)=\frac{x^4 - 1}{x - 1}=\frac{(x^2 + 1)(x + 1)(x - 1)}{x - 1}\) (for \(x
eq1\)).

Step2: Cancel the common factor

Since \(x
eq1\) (because the original function is undefined at \(x = 1\)), we can cancel out the common factor \((x - 1)\) from the numerator and the denominator. After canceling, we get \(f(x)=x^3+x^2+x + 1\) (for \(x
eq1\)). To make the discontinuity removable, we define \(g(x)\) as this simplified function, and since the limit as \(x\) approaches \(1\) of \(f(x)\) exists (we can plug in \(x = 1\) into the simplified function to find the limit), the discontinuity at \(x = 1\) is removable, and \(g(x)=x^3+x^2+x + 1\) (for all \(x\) where the function is defined, including \(x = 1\) now).

Answer:

\(g(x)=x^3 + x^2 + x + 1\)