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Step-by-Step Solution
Step 1: Identify the derivative of the polynomial
Given that the polynomial $f(x)$ is of degree 4 and has local extremes at $x = -1, 0, 1$. At these points, the derivative must be zero. Thus,
$f'(x) = 0 \text{ at } x = -1, \, 0, \, 1.$
Since $f'(x)$ is a cubic polynomial (because $f(x)$ is a quartic polynomial), we can write
$ f'(x) = k(x+1)x(x-1), $
for some constant $k \neq 0.$
Step 2: Integrate to find the general form of $f(x)$
We integrate $f'(x)$ to find $f(x)$:
\[
f(x) \;=\; \int k (x+1)x(x-1)\,dx.
\]
Carrying out the integration term-by-term,
\[
f(x) = k \left[ \frac{x^4}{4} - \frac{x^2}{2} \right] + C,
\]
where $C$ is the constant of integration.
Step 3: Use the condition $f(x) = f(0)$
We know $f(0) = C.$ The set
\[
S = \{x \in \mathbb{R} \mid f(x) = f(0)\}
\]
implies
\[
k \left[ \frac{x^4}{4} - \frac{x^2}{2} \right] + C \;=\; C.
\]
Subtracting $C$ from both sides gives:
\[
k \left[ \frac{x^4}{4} - \frac{x^2}{2} \right] = 0.
\]
Since $k \neq 0,$ we solve
\[
\frac{x^4}{4} - \frac{x^2}{2} = 0.
\]
Step 4: Factor and find the roots
Rewrite and factor:
\[
\frac{x^4}{4} - \frac{x^2}{2} = \frac{x^2}{4}\left(x^2 - 2\right) = 0.
\]
This expression is zero if
\[
x^2 = 0 \quad \text{or} \quad x^2 = 2.
\]
Hence, the solutions are
\[
x = 0, \quad x = \sqrt{2}, \quad x = -\sqrt{2}.
\]
Step 5: Interpret the result
The set $S$ has the three distinct real values:
\[
\{\, 0,\; -\sqrt{2},\; \sqrt{2} \}.
\]
Among these, $0$ is rational, while $-\sqrt{2}$ and $\sqrt{2}$ are irrational. Thus, there are exactly two irrational and one rational number in the set $S.$
Answer:
The set $S$ contains exactly two irrational numbers and one rational number.
