Answer:
x = -3/2
Step-by-step explanation:
The removable discontinuity is at the value of x where a denominator factor cancels a numerator factor, and those factors are zero.
[tex]p(x)=\dfrac{2x+3}{4x^2-9}=\dfrac{2x+3}{(2x+3)(2x-3)}=\dfrac{2x+3}{2x+3}\cdot\dfrac{1}{2x-3}\\\\p(x)=\dfrac{1}{2x-3}\qquad x\ne-3/2[/tex]
That cancellation occurs where 2x+3 = 0, at x=-3/2.
The function p(x) has a removable discontinuity when x = -3/2.
