To solve for [tex]\( f(x) = \frac{2}{x+2} \)[/tex] at given values of [tex]\( x \)[/tex], let's evaluate the function step-by-step for specific values of [tex]\( x \)[/tex]: 1, 3, -1, 4, and -2.
1. Evaluate [tex]\( f(1) \)[/tex]:
Substitute [tex]\( x = 1 \)[/tex] into the function.
[tex]\[
f(1) = \frac{2}{1 + 2} = \frac{2}{3} \approx 0.6666666666666666
\][/tex]
2. Evaluate [tex]\( f(3) \)[/tex]:
Substitute [tex]\( x = 3 \)[/tex] into the function.
[tex]\[
f(3) = \frac{2}{3 + 2} = \frac{2}{5} = 0.4
\][/tex]
3. Evaluate [tex]\( f(-1) \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into the function.
[tex]\[
f(-1) = \frac{2}{-1 + 2} = \frac{2}{1} = 2.0
\][/tex]
4. Evaluate [tex]\( f(4) \)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] into the function.
[tex]\[
f(4) = \frac{2}{4 + 2} = \frac{2}{6} = \frac{1}{3} \approx 0.3333333333333333
\][/tex]
5. Evaluate [tex]\( f(-2) \)[/tex]:
The function is undefined for [tex]\( x = -2 \)[/tex], since the denominator would become zero:
[tex]\[
f(-2) = \frac{2}{-2 + 2} = \frac{2}{0} \quad (\text{Undefined})
\][/tex]
Therefore, [tex]\( f(-2) \)[/tex] does not exist, or we can denote it as [tex]\( \text{None} \)[/tex].
So, summarizing these evaluations:
- [tex]\( f(1) \approx 0.6666666666666666 \)[/tex]
- [tex]\( f(3) = 0.4 \)[/tex]
- [tex]\( f(-1) = 2.0 \)[/tex]
- [tex]\( f(4) \approx 0.3333333333333333 \)[/tex]
- [tex]\( f(-2) \)[/tex] is undefined or [tex]\( \text{None} \)[/tex]
Thus, the results for the function [tex]\( f(x) = \frac{2}{x+2} \)[/tex] at the specified values of [tex]\( x \)[/tex] are:
[tex]\[
(0.6666666666666666, 0.4, 2.0, 0.3333333333333333, \text{None})
\][/tex]
