Question

(b) ( f(x) = \frac{x^3 - x^2 - 56x}{x - 8}, quad a = 8 )
( \bigcirc ) the discontinuity is removable.
( \bigcirc ) the discontinuity is not removable.
( g(x) = square )

Explanation:

Step1: Factor the numerator

First, factor the numerator \(x^3 - x^2 - 56x\). We can factor out an \(x\) first:
\(x^3 - x^2 - 56x = x(x^2 - x - 56)\)
Then, factor the quadratic \(x^2 - x - 56\). We need two numbers that multiply to \(-56\) and add to \(-1\). Those numbers are \(-8\) and \(7\), so:
\(x^2 - x - 56 = (x - 8)(x + 7)\)
So the numerator factors to \(x(x - 8)(x + 7)\).

Step2: Simplify the function \(f(x)\)

Now, the function \(f(x)=\frac{x^3 - x^2 - 56x}{x - 8}\) can be rewritten with the factored numerator:
\(f(x)=\frac{x(x - 8)(x + 7)}{x - 8}\)
For \(x
eq 8\), we can cancel out the \((x - 8)\) terms:
\(f(x)=x(x + 7)=x^2 + 7x\) (for \(x
eq 8\))

Step3: Determine the type of discontinuity and find \(g(x)\)

Since we were able to cancel the factor that makes the denominator zero (\(x - 8\)), the discontinuity at \(x = 8\) is removable. The function \(g(x)\) that removes the discontinuity is the simplified form of \(f(x)\) (since the discontinuity is removable, \(g(x)\) should be equal to \(f(x)\) for all \(x\) except at the point of discontinuity, and defined at \(x = 8\) to make it continuous). So \(g(x)=x^2 + 7x\) (and we can check that at \(x = 8\), \(g(8)=8^2 + 7\times8=64 + 56 = 120\), and the limit of \(f(x)\) as \(x\) approaches \(8\) is also \(120\), so this removes the discontinuity).

Answer:

The discontinuity is removable.
\(g(x)=\boxed{x^2 + 7x}\)