Evaluar la función $$ f(x)=\frac{3-x}{-16+x} $$ en los valores indicados: $$ f(2)=\square $$ $$ f(-2)=\square $$ $$ f\left(\frac{1}{2}\right)=\square $$ $$ f(a)=\square $$ $$ f(a-1)=\square $$

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1. Para $$ x=2 $$:

$$
   f(2)=\frac{3-2}{-16+2}=\frac{1}{-14}=-\frac{1}{14}
   $$

2. Para $$ x=-2 $$:

$$
   f(-2)=\frac{3-(-2)}{-16+(-2)}=\frac{3+2}{-18}=\frac{5}{-18}=-\frac{5}{18}
   $$

3. Para $$ x=\frac{1}{2} $$:

$$
   f\left(\frac{1}{2}\right)=\frac{3-\frac{1}{2}}{-16+\frac{1}{2}}=\frac{\frac{6-1}{2}}{-\frac{32-1}{2}}=\frac{\frac{5}{2}}{-\frac{31}{2}}=-\frac{5}{31}
   $$

4. Para $$ x=a $$:

$$
   f(a)=\frac{3-a}{-16+a}=\frac{3-a}{a-16}
   $$

5. Para $$ x=a-1 $$:

$$
   f(a-1)=\frac{3-(a-1)}{-16+(a-1)}=\frac{3-a+1}{a-17}=\frac{4-a}{a-17}
   $$
1. $$ f(2) = -\frac{1}{14} $$ 2. $$ f(-2) = -\frac{5}{18} $$ 3. $$ f\left(\frac{1}{2}\right) = -\frac{5}{31} $$ 4. $$ f(a) = \frac{3 - a}{a - 16} $$ 5. $$ f(a - 1) = \frac{4 - a}{a - 17} $$

Evaluar la función \( f(x)=\frac{3-x}{-16+x} \) en los valores indicados: \( f(2)=\square \) \( f(-2)=\square \) \( f\left(\frac{1}{2}\right)=\square \) \( f(a)=\square \) \( f(a-1)=\square \)

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