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Step-by-Step Solution
Step 1: Understand the Function f(x)
We are given a function
$ f(x) = x^2 $,
where $ x \in \mathbb{R} $. This function takes any real number $x$ and maps it to $x^2$.
Step 2: Define g(A)
For any subset
$ A \subseteq \mathbb{R} $,
the set
$ g(A) $
is defined as
$ g(A) = \{\, x \in \mathbb{R} : f(x) \in A \}. $
In other words,
$ g(A) $
is the pre-image (or inverse image) of the set
$A$
under the function
$f$.
That is,
$x$
belongs to
$g(A)$
if and only if
$f(x)$
belongs to
$A$.
Step 3: Identify the Set S
We are given
$ S = [0,4]. $
This is the closed interval from 0 to 4.
Step 4: Find g(S)
By definition,
$ g(S) = \{\, x \in \mathbb{R} : f(x) \in S \} = \{\, x \in \mathbb{R} : x^2 \in [0,4] \}. $
The values of $x$ satisfying
$ 0 \le x^2 \le 4 $
are precisely those in the interval
$ -2 \le x \le 2 .$
Hence,
$ g(S) = [-2, 2]. $
Step 5: Compute f(g(S))
Next, we apply $f$ to all elements of $g(S)$. Since
$ g(S) = [-2,2], $
$ f(g(S)) = \{\, x^2 : x \in [-2,2] \} = [0,4].
$
Notice that
$ [0,4] $
is the original set
$S$.
Thus,
$ f(g(S)) = S.
$
Step 6: Compare with f(S)
Now, consider
$ f(S). $
Since
$ S = [0,4], $
when we apply $f(x)=x^2$ to every $x$ in $[0,4]$, we get
$ f(S) = \{\, x^2 : x \in [0,4] \} = [0,16].
$
Therefore,
$ f(g(S)) = [0,4] $
while
$ f(S) = [0,16]. $
Hence
$ f(g(S)) \ne f(S).
Step 7: Find g(f(S))
Consider
$ f(S) = [0,16]. $
Then
$ g(f(S)) = \{\, x \in \mathbb{R} : x^2 \in [0,16] \} = \{\, x : -4 \le x \le 4 \} = [-4,4].
$
We observe that
$ g(f(S)) = [-4,4], $
while we have
$ g(S) = [-2,2].
$
Therefore,
$ g(f(S)) \ne g(S).
Step 8: Identify the Statement That Is Not True
The statement
"g(f(S)) = g(S)"
is clearly not true because we found
$ g(f(S)) = [-4,4] $
whereas
$ g(S) = [-2,2].
$
Therefore, the statement
g(f(S)) = g(S)
is the one that is not true.
Conclusion
The correct answer, that is, the statement that fails to hold, is
g(f(S)) = g(S).
