Answer:
Evaluating the function is just replacing the "x" (or the "t" in the second case) by the value that we want:
1) q(x) = 1/(x^2 - 9)
a) q(0) = 1/(0^2 - 9) = 1/(-9) = - 1/9
b) q(3) = 1/(3^2 - 9) = 1/(9 - 9) = 1/0 = indetermined, we could think this as:
[tex]\lim_{x \to \ 3} q(x) = infinity[/tex]
c) q(y + 3) = 1/( (y + 3)^2 - 9) = 1/(y^2 + 6*y + 9 - 9) = 1/(y^2 + 6*y)
2) q(t) = (2*t^2 + 3)/t^2
a) q(2) = (2*2^2 + 3)/2^2 = (2*4 + 3)/4 = 11/4
b) q(0) = (2*0^2 + 3)/0^2 = 3/0
Again, we can not divide by zero, this is simmilar to the case 1.b
c) q(-x) = (2*(-x)^2 + 3)/(-x)^2 = (2*x^2 + 3)/x^2
