Question

find the x - values (if any) at which f is not continuous. if enter dne.)
f(x)=\frac{x + 4}{x^{2}-2x - 24}
removable discontinuities
x =
nonremovable discontinuities
x =

Explanation:

Step1: Factor denominator

Denominator \(x^2 - 2x - 24\) factors as \((x - 6)(x + 4)\).

Step2: Find denominator roots

Set \((x - 6)(x + 4) = 0\), roots \(x = 6\) and \(x = -4\).

Step3: Identify removable discontinuity

Numerator has \((x + 4)\), cancels with denominator's \((x + 4)\). So \(x = -4\) is removable.

Step4: Identify nonremovable discontinuity

\(x = 6\) remains in denominator, nonremovable.

Answer:

removable discontinuities \(x = -4\)
nonremovable discontinuities \(x = 6\)