If the equation a_n xⁿ + a_{n-1} x^{n-1} + ... + a₁ x = 0, a₁ ≠ 0, n ≥ 2, has a positive root x = α,

Q: Wait, why can we assume $f(0) = 0$? The problem doesn't explicitly state it.

That is a fantastic observation! Look closely at the structure of the polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$. Every single term in this expression has an $x$ attached to it. When you substitute $x=0$, every term becomes zero. Therefore, $f(0)$ is mathematically guaranteed to be $0$. It is a hidden property of the polynomial that we must uncover to solve the problem.

Q: Can we use the Mean Value Theorem instead of Rolle's Theorem?

You absolutely can! Rolle's Theorem is actually a special case of the Mean Value Theorem (MVT). MVT states that there exists a $c$ such that $f'(c) = \frac{f(\alpha) - f(0)}{\alpha - 0}$. Since $f(\alpha) = 0$ and $f(0) = 0$, the right side becomes $\frac{0-0}{\alpha} = 0$. So, $f'(c) = 0$. Rolle's Theorem is just the more direct tool for this specific scenario, but your intuition to use MVT is mathematically sound and shows you understand the broader calculus landscape.

Q: Does this result hold true regardless of the degree $n$?

Yes, it does! The beauty of Rolle's Theorem is that it depends only on the continuity and differentiability of the function, not on its degree. As long as $n \geq 2$, the polynomial will behave smoothly, and the existence of a root for the derivative between the roots of the original function is guaranteed. This is the power of general theorems in mathematics—they work for any $n$!

Question

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$, $a_1 
eq 0, n \geq 2$, has a positive root $x = \alpha$, then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root, which is

Options:

(A) greater than $\alpha$

(B) smaller than $\alpha$

(C) greater than or equal to $\alpha$

(D) equal to $\alpha$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, rolling landscape defined by a polynomial function:

$$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$$

At first glance, this expression might look like a chaotic jumble of coefficients and powers. However, there is a secret hidden in plain sight.

Because every term in this polynomial contains at least one factor of $x$, we know that when $x=0$, the entire function collapses to zero. That is, $f(0) = 0$. This means our curve is anchored at the origin.

The problem states there is a positive root, $\alpha$. This means that at some point further down the x-axis, the curve dips back down and kisses the axis again at $f(\alpha) = 0$.

### The Calculus Bridge

Now, consider the second equation:

$$n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$$

If you have spent enough time with calculus, your eyes should light up. This is not just a random equation; it is the derivative of our original function, $f'(x) = 0$.

By applying the power rule, $\frac{d}{dx}(x^n) = n x^{n-1}$, to each term of $f(x)$, we arrive exactly at this second expression. Geometrically, finding a root of the derivative means finding a point where the slope of the tangent line is zero—a peak or a valley in our landscape.

### The Power of Rolle's Theorem

This is where the magic happens. We have a function $f(x)$ that is continuous and differentiable (because all polynomials are) on the interval $[0, \alpha]$.

We know $f(0) = 0$ and $f(\alpha) = 0$. These are the exact conditions required for **Rolle's Theorem**.

Rolle's Theorem is like a promise from the universe of mathematics: if you start at zero and end at zero, and you travel along a smooth path, you must have turned around at least once. Specifically, there must exist some point $c$ strictly between $0$ and $\alpha$ where the slope is zero, meaning $f'(c) = 0$.

### The Final Revelation

Since $c$ is guaranteed to be in the open interval $(0, \alpha)$, we have found our positive root for the second equation. It is not just any root; it is a root that is **strictly smaller than $\alpha$**.

We have successfully navigated from the algebraic definition of a polynomial to the geometric reality of its derivative, using the elegant bridge of Rolle's Theorem.

The next time you see a complex polynomial, don't be intimidated by the powers of $n$. Look for the roots, look for the derivative, and remember that the geometry of the curve will always guide you to the answer.

Select Answer:

Visualized Solution

Defining the Function $f (x)$

Properties of Polynomials

Evaluating $f (0)$

The Given Positive Root $α$

Analyzing the Second Equation

Differentiating $f (x)$

Setting up Rolle's Theorem

Verifying Rolle's Conditions

Applying Rolle's Theorem

The Horizontal Tangent

Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Calculus Bridge

The Power of Rolle's Theorem

The Final Revelation

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