Animated Solution for Mathematics - Differentiation: If $f(x)$ is a non-zero polynomial of degree four, having local extreme points at $x=-1,0,1$; then the set $S=\{x\in \mathbb{R}:f(x)=f(0)\}$ contains exactly :

Imagine you are standing on a landscape defined by a fourth-degree polynomial. It is a rolling, elegant curve, a 'W' shape that dips and rises.

The problem states that this landscape has local extreme points—peaks and valleys—at $x=-1$, $x=0$, and $x=1$. In the language of calculus, this is our golden ticket.

At every local peak or valley, the slope of the tangent line is perfectly horizontal. This means the first derivative, $f^{\prime }(x)$, must be zero at these three specific coordinates. This is the heartbeat of our problem.

Since $f(x)$ is a polynomial of degree four, its derivative $f^{\prime }(x)$ must be a cubic polynomial (degree three). We know its roots are $-1,0,$ and $1$.

By the Factor Theorem, we can immediately write the derivative as $f^{\prime }(x)=k(x+1)(x)(x-1)$, where $k$ is a non-zero constant. This $k$ is the 'DNA' of our curve, determining its steepness.

Let's simplify this expression using the difference of squares identity: $(x+1)(x-1)=x^2-1$. Multiplying this by the remaining $x$, we get:

Now, we need to reverse the process. To find $f(x)$, we integrate $f^{\prime }(x)$ with respect to $x$.

Integrating $k(x^3-x)$ gives us:

That constant $C$ is the vertical shift of our graph—the y-intercept. The problem asks us to analyze the set $S=\{x\in \mathbb{R}:f(x)=f(0)\}$.

To do this, we first need to evaluate $f(0)$. Substituting $x=0$ into our expression, the terms involving $x$ vanish, leaving us with $f(0)=C$.

Now, we set $f(x)=f(0)$. Substituting our expressions, we get:

The constant $C$ appears on both sides, and they cancel out perfectly. We are left with:

Since $k eq 0$, we can divide by $k$ and multiply by $4$ to clear the fractions, resulting in $x^4-2x^2=0$. Factoring this, we get $x^2(x^2-2)=0$.

This yields three distinct solutions: $x=0$, $x=\sqrt{2}$, and $x=-\sqrt{2}$.

We have our set $S=\{0,\sqrt{2},-\sqrt{2}\}$.

Now, let's classify these numbers. The number $0$ is rational, while the numbers $\sqrt{2}$ and $-\sqrt{2}$ are irrational.

Thus, our set contains exactly two irrational numbers and one rational number. We have navigated the calculus, mastered the algebra, and arrived at the truth.