Answer :
Let's analyze the functions [tex]\( f(x) = 3x^3 + 2 \)[/tex] and [tex]\( g(x) = \sqrt[3]{\frac{x-2}{3}} \)[/tex] to determine whether any of the statements I, II, or III are true.
1. First, we need to check if [tex]\( f(g(x)) = x \)[/tex] for all real [tex]\( x \)[/tex].
[tex]\[ f(g(x)) = f\left(\sqrt[3]{\frac{x-2}{3}}\right) \][/tex]
Now, let's substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\sqrt[3]{\frac{x-2}{3}}\right) = 3\left(\sqrt[3]{\frac{x-2}{3}}\right)^3 + 2 \][/tex]
Simplifying inside the brackets:
[tex]\[ 3 \left(\frac{x-2}{3}\right) + 2 = x - 2 + 2 = x \][/tex]
Therefore, [tex]\( f(g(x)) = x \)[/tex] is true.
2. Next, let's verify if [tex]\( g(f(x)) = x \)[/tex] for all real [tex]\( x \)[/tex].
[tex]\[ g(f(x)) = g(3x^3 + 2) \][/tex]
Now, substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(3x^3 + 2) = \sqrt[3]{\frac{3x^3 + 2 - 2}{3}} = \sqrt[3]{x^3} = x \][/tex]
Therefore, [tex]\( g(f(x)) = x \)[/tex] is also true.
3. Finally, we check if the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverse functions. For this, both [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex] must hold true.
Since we have shown that both [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex] for all [tex]\( x \)[/tex], [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are indeed inverse functions.
Given these findings:
- I: [tex]\( f(g(x)) = x \)[/tex] is true for all real [tex]\( x \)[/tex].
- II: [tex]\( g(f(x)) = x \)[/tex] is true for all real [tex]\( x \)[/tex].
- III: Functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverse functions.
Thus, the correct answer is:
C. I, II, and III are true.
1. First, we need to check if [tex]\( f(g(x)) = x \)[/tex] for all real [tex]\( x \)[/tex].
[tex]\[ f(g(x)) = f\left(\sqrt[3]{\frac{x-2}{3}}\right) \][/tex]
Now, let's substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\sqrt[3]{\frac{x-2}{3}}\right) = 3\left(\sqrt[3]{\frac{x-2}{3}}\right)^3 + 2 \][/tex]
Simplifying inside the brackets:
[tex]\[ 3 \left(\frac{x-2}{3}\right) + 2 = x - 2 + 2 = x \][/tex]
Therefore, [tex]\( f(g(x)) = x \)[/tex] is true.
2. Next, let's verify if [tex]\( g(f(x)) = x \)[/tex] for all real [tex]\( x \)[/tex].
[tex]\[ g(f(x)) = g(3x^3 + 2) \][/tex]
Now, substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(3x^3 + 2) = \sqrt[3]{\frac{3x^3 + 2 - 2}{3}} = \sqrt[3]{x^3} = x \][/tex]
Therefore, [tex]\( g(f(x)) = x \)[/tex] is also true.
3. Finally, we check if the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverse functions. For this, both [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex] must hold true.
Since we have shown that both [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex] for all [tex]\( x \)[/tex], [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are indeed inverse functions.
Given these findings:
- I: [tex]\( f(g(x)) = x \)[/tex] is true for all real [tex]\( x \)[/tex].
- II: [tex]\( g(f(x)) = x \)[/tex] is true for all real [tex]\( x \)[/tex].
- III: Functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverse functions.
Thus, the correct answer is:
C. I, II, and III are true.
