a)
f(-3), f(1) i f(3).
f(x)=\frac{6x}{x^2+9}
x^2+9>0
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D:x\in \mathbb R
f(-3)=\frac{6*(-3)}{(-3)^2+9}=\frac{-18}{9+9}=\frac{-18}{18}=-1
f(1)=\frac{6*1}{1^2+9}=\frac{6}{10}=\frac{3}{5}
f(3)=\frac{6*3}{3^2+9}=\frac{18}{18}=1
b)
f(x)=\frac{x^2+2x}{x^2-4}
x^2-4\ne0
(x-2)(x+2)\ne0
x-2\ne0 i x+2\ne0
x\ne 2 i x\ne -2 , D=\mathbb R\backslash \{-2,2\}
f(-3)=\frac{(-3)^2+2*(-3)}{(-3)^2-4}=\frac{9-6}{9-4}=\frac{3}{5}
f(1)=\frac{1^2+2*1}{1^2-4}=\frac{3}{-3}=-1
f(3)=\frac{3^2+2*3}{3^2-4}=\frac{9+6}{5}=\frac{15}{5}=3
c)
f(x)=\frac{x^2+21}{x^2-3}
x^2-3\ne0
(x-\sqrt3)(x+\sqrt3)\ne0
x-\sqrt3\ne 0 i x+\sqrt3\ne 0
x\ne \sqrt3 i x\ne -\sqrt3, D=\mathbb R\backslash \{-\sqrt3,\sqrt3\}
f(-3)=\frac{(-3)^2+21}{(-3)^2-3}=\frac{9+21}{9-3}=\frac{30}{6}=5
f(1)=\frac{1^2+21}{1^2-3}=\frac{22}{-2}=-11
f(3)=\frac{3^2+21}{3^2-3}=\frac{9+21}{9-3}=\frac{30}{6}=5