Question

question 8 of 10
which function has a removable discontinuity?

a. ( p(x) = \frac{x - 2}{x^2 - x - 2} )
b. ( g(x) = \frac{2x - 1}{x} )
c. ( f(x) = \frac{5x}{1 - x^2} )
d. ( h(x) = \frac{x^2 - x + 2}{x + 1} )

Explanation:

Step1: Recall removable discontinuity

A function has a removable discontinuity if the limit exists at a point where the function is undefined, and this happens when a common factor can be canceled from the numerator and denominator.

Step2: Analyze Option A

Factor the denominator of \( p(x)=\frac{x - 2}{x^{2}-x - 2} \). We factor \( x^{2}-x - 2 \) as \( (x - 2)(x+1) \) (since \( -2=(-2)\times1 \) and \( -2 + 1=-1 \)). So \( p(x)=\frac{x - 2}{(x - 2)(x + 1)} \), for \( x
eq2 \), we can cancel the \( (x - 2) \) terms, getting \( p(x)=\frac{1}{x + 1} \) (with a hole at \( x = 2 \)).

Step3: Analyze Option B

For \( g(x)=\frac{2x-1}{x} \), the denominator is zero when \( x = 0 \), and the numerator at \( x = 0 \) is \( - 1
eq0 \), so it has a non - removable (infinite) discontinuity at \( x = 0 \).

Step4: Analyze Option C

For \( f(x)=\frac{5x}{1 - x^{2}}=\frac{5x}{(1 - x)(1 + x)} \), the denominator is zero at \( x=\pm1 \), and the numerator at \( x = 1 \) is \( 5
eq0 \) and at \( x=-1 \) is \( - 5
eq0 \), so it has non - removable discontinuities at \( x=\pm1 \).

Step5: Analyze Option D

For \( h(x)=\frac{x^{2}-x + 2}{x + 1} \), we can try to factor the numerator. The discriminant of the numerator \( x^{2}-x + 2 \) is \( \Delta=(-1)^{2}-4\times1\times2=1 - 8=-7<0 \), so the numerator has no real roots. The denominator is zero at \( x=-1 \), and the numerator at \( x=-1 \) is \( (-1)^{2}-(-1)+2=1 + 1+2 = 4
eq0 \), so it has a non - removable (infinite) discontinuity at \( x=-1 \).

Answer:

A. \( p(x)=\frac{x - 2}{x^{2}-x - 2} \)