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Step-by-Step Solution
Step 1: Express the derivative of the polynomial
Given that the polynomial function $f(x)$ is of degree 4 and has critical points at $x = -1, 0,$ and $1$. A critical point occurs where the derivative $f'(x)$ is zero. Hence, we can write:
$f'(x) = a(x - 1)(x + 1)x \quad (\text{for some constant } a).$
Step 2: Integrate to find the polynomial $f(x)$
Integrate $f'(x)$ term by term with respect to $x$ to determine $f(x)$:
\[
f'(x) = a \, x(x^2 - 1) = a \left(x^3 - x\right).
\]
\[
f(x) = \int a \left(x^3 - x\right) \, dx = a \left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C,
\]
where $C$ is the constant of integration.
Step 3: Apply the condition $f(x) = f(0)$
We are told to find the set $T = \{ x \in \mathbb{R} \mid f(x) = f(0) \}$. First, compute $f(0)$:
\[
f(0) = a \left(\frac{0^4}{4} - \frac{0^2}{2}\right) + C = C.
\]
Setting $f(x) = f(0)$ gives
\[
a \left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C = C.
\]
Subtract $C$ from both sides:
\[
a \left(\frac{x^4}{4} - \frac{x^2}{2}\right) = 0.
\]
Step 4: Solve for $x$
Factor the expression inside the parentheses:
\[
\frac{a}{4} \, x^2(x^2 - 2) = 0.
\]
This equation is satisfied if $x^2 = 0$ or $x^2 - 2 = 0$. Therefore, the solutions are:
$x = 0$
$x = \sqrt{2}$
$x = -\sqrt{2}$
Hence, the set $T$ is $\{\,0, \sqrt{2}, -\sqrt{2}\}$.
Step 5: Compute the sum of the squares of the elements of $T$
The elements of $T$ are $0$, $\sqrt{2}$, and $-\sqrt{2}$. Calculate their squares and sum them:
\[
0^2 + \left(\sqrt{2}\right)^2 + \left(-\sqrt{2}\right)^2 = 0 + 2 + 2 = 4.
\]
Therefore, the required sum of squares of all elements of $T$ is $\boxed{4}$.
