Suppose f(x) is a polynomial of degree four, having critical points at -1,0,1. If T = {x ∈ R | f(x)

Q: Why does the constant $C$ disappear? Does it mean the function is not unique?

Great observation! The constant $C$ represents a vertical shift of the entire polynomial. Since the condition $f(x) = f(0)$ is essentially asking for the intersection of the curve with a horizontal line at height $f(0)$, shifting the entire graph up or down by $C$ does not change the $x$-coordinates where this intersection occurs. The shape of the curve is determined by $A$, but the horizontal positions of the intersection points are independent of $C$.

Q: Can we use the property of roots of polynomials directly instead of solving for $x$?

You could, but in this case, solving for $x$ is quite straightforward. We get $x^2(x^2 - 2) = 0$, which gives us $x = 0$ (a double root) and $x = \pm\sqrt{2}$. Since the set $T$ only cares about the distinct values of $x$ that satisfy the equation, we list them as $\{0, \sqrt{2}, -\sqrt{2}\}$. Solving directly is often less prone to error than using Vieta's formulas when the roots are this simple.

Q: Is it possible for $f(x)$ to have other critical points?

No. We are given that $f(x)$ is a degree four polynomial, which means $f'(x)$ is a degree three polynomial. A degree three polynomial can have at most three roots. Since we have already identified three distinct critical points ($-1, 0, 1$), these must be the only critical points of the function.

Question

Suppose $f(x)$ is a polynomial of degree four, having critical points at $-1,0,1$. If $T = \{x \in \mathbb{R} | f(x) = f(0)\}$, then the sum of squares of all the elements of $T$ is :

Options:

(A) $4$

(B) $2$

(C) $6$

(D) $8$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a landscape defined by a fourth-degree polynomial. It is a smooth, rolling terrain with peaks and valleys.

In calculus, we focus on the 'critical points'—the exact locations where the landscape levels off, where the slope of the ground is perfectly horizontal. These are the points where the derivative, $f'(x)$, is zero.

In our problem, we are given these critical points at $x = -1$, $x = 0$, and $x = 1$. This is the DNA of our function.

Because $f(x)$ is a polynomial of degree four, its derivative $f'(x)$ must be a cubic polynomial. Since we know the roots are $-1, 0,$ and $1$, we can write the derivative as:

$$f'(x) = A(x+1)(x)(x-1)$$

Here, $A$ is a non-zero constant that dictates the 'stretch' or 'steepness' of our curve.

### From Slope to Height: The Integration Journey

Now, we have the slope, but we want the height. To move from the derivative $f'(x)$ back to the original function $f(x)$, we must perform the inverse operation: integration.

Let us expand our derivative first to make the integration easier. Using the difference of squares, $(x+1)(x-1)$ becomes $(x^2 - 1)$. Multiplying this by $x$, we get:

$$f'(x) = A(x^3 - x)$$

Now, we integrate: $f(x) = \int A(x^3 - x) dx$. Applying the power rule, where the integral of $x^n$ is $\frac{x^{n+1}}{n+1}$, we find:

$$f(x) = A\left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C$$

That constant $C$ is crucial—it represents the vertical position of our landscape. Without it, we would only know the shape, not the height.

### The Intersection: Solving for the Set $T$

The problem asks us to find the set $T = \{x \in \mathbb{R} \mid f(x) = f(0)\}$. Geometrically, this is asking: "At what other points does the function reach the same height as it does at $x = 0$?"

First, let us find $f(0)$. By substituting $x = 0$ into our expression for $f(x)$, the terms involving $x$ vanish, leaving us with $f(0) = C$.

Now, we set $f(x) = f(0)$, which means:

$$A\left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C = C$$

Notice the elegance here! The constant $C$ cancels out perfectly from both sides. We are left with $A\left(\frac{x^4}{4} - \frac{x^2}{2}\right) = 0$.

Since $A$ is non-zero, we can divide it away, leaving us with the biquadratic equation:

$$\frac{x^4}{4} - \frac{x^2}{2} = 0$$

### The Final Calculation

To solve $\frac{x^4}{4} - \frac{x^2}{2} = 0$, let us multiply by 4 to clear the fractions:

$$x^4 - 2x^2 = 0$$

Factoring out $x^2$, we get $x^2(x^2 - 2) = 0$. This gives us the roots $x = 0$ and $x^2 = 2$, which means $x = \pm\sqrt{2}$.

The elements of our set $T$ are $\{0, \sqrt{2}, -\sqrt{2}\}$. The question asks for the sum of the squares of these elements.

Calculating this, we get:

$$(0)^2 + (\sqrt{2})^2 + (-\sqrt{2})^2 = 0 + 2 + 2 = 4$$

Through the power of calculus and algebraic symmetry, we have navigated the landscape of this polynomial and arrived at our destination: **4**.

Animated Solution for Mathematics - Differentiation: Suppose $f (x)$ is a polynomial of degree four, having critical points at $- 1, 0, 1$ . If $T = {x \in R ∣ f (x) = f (0)}$ , then the sum of squares of all the elements of $T$ is :

Select Answer:

Visualized Solution

Identify the Critical Points

Formulate the Derivative $f^{'} (x)$

Expand the Derivative Expression

Integrate to find $f (x)$

Perform the Integration

Define the Set $T$

Calculate $f (0)$

Set up the Equation $f (x) = f (0)$

Simplify the Equation

Solve for $x$

Find the Elements of Set $T$

Calculate the Sum of Squares

The Sigma Insight: Maxima and Minima

Analyzing the Setup

From Slope to Height

The Intersection

The Final Calculation

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Select Answer:

Visualized Solution

Identify the Critical Points

Formulate the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Expand the Derivative Expression

Integrate to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Perform the Integration

Define the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

Set up the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=f(0)

Simplify the Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Find the Elements of Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Calculate the Sum of Squares

The Sigma Insight: Maxima and Minima

Analyzing the Setup

From Slope to Height

The Intersection

The Final Calculation

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Let $f (x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$ . If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is equal to

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Let $f : R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

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Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

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