Let's analyze the statements using the given functions [tex]\( f(x) = -2x - 1 \)[/tex] and [tex]\( g(x) = -\frac{1}{2}x + \frac{1}{2} \)[/tex] step-by-step to determine their validity.
### Checking [tex]\( f(g(x)) \)[/tex]
First, we need to find [tex]\( f(g(x)) \)[/tex].
1. Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(g(x)) = f\left(-\frac{1}{2}x + \frac{1}{2}\right)
\][/tex]
2. Replace the argument of [tex]\( f \)[/tex] with [tex]\(-\frac{1}{2}x + \frac{1}{2}\)[/tex]:
[tex]\[
f\left(-\frac{1}{2}x + \frac{1}{2}\right) = -2\left(-\frac{1}{2}x + \frac{1}{2}\right) - 1
\][/tex]
3. Distribute and simplify:
[tex]\[
-2\left(-\frac{1}{2}x + \frac{1}{2}\right) - 1 = (-2) \cdot \left(-\frac{1}{2}x\right) + (-2) \cdot \left(\frac{1}{2}\right) - 1
\][/tex]
[tex]\[
= x - 1 - 1
\][/tex]
[tex]\[
= x - 2
\][/tex]
So, [tex]\( f(g(x)) = x - 2 \neq x \)[/tex].
Therefore, statement I is not true.
### Checking [tex]\( g(f(x)) \)[/tex]
Next, we need to find [tex]\( g(f(x)) \)[/tex].
1. Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[
g(f(x)) = g(-2x - 1)
\][/tex]
2. Replace the argument of [tex]\( g \)[/tex] with [tex]\(-2x - 1\)[/tex]:
[tex]\[
g(-2x - 1) = -\frac{1}{2}(-2x - 1) + \frac{1}{2}
\][/tex]
3. Distribute and simplify:
[tex]\[
-\frac{1}{2}(-2x - 1) + \frac{1}{2} = -\frac{1}{2} \cdot (-2x) - \frac{1}{2} \cdot (1) + \frac{1}{2}
\][/tex]
[tex]\[
= x + \frac{1}{2} - \frac{1}{2}
\][/tex]
[tex]\[
= x
\][/tex]
So, [tex]\( g(f(x)) = x \)[/tex].
Therefore, statement II is true.
### Checking if [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverse functions
To determine if [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverse functions, both [tex]\( f(g(x)) \)[/tex] must equal [tex]\( x \)[/tex] and [tex]\( g(f(x)) \)[/tex] must equal [tex]\( x \)[/tex].
Since [tex]\( f(g(x)) = x - 2 \neq x \)[/tex] fails, [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses.
Therefore, statement III is not true.
### Conclusion
Only statement II is true.
The correct answer is:
B. II only
