Question

Someone please help me

Answer 1

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.

  $p(x)=\dfrac{2x+3}{4x^2-9}=\dfrac{2x+3}{(2x+3)(2x-3)}=\dfrac{2x+3}{2x+3}\cdot\dfrac{1}{2x-3}\\\\p(x)=\dfrac{1}{2x-3}\qquad x\ne-3/2$

That cancellation occurs where 2x+3 = 0, at x=-3/2.

The function p(x) has a removable discontinuity when x = -3/2.