求助：不理解Fermat's Theorem与Rolle's Theorem，如何证明多项式零点个数？

Hey there! No worries at all—we all start out confused with these theorems, and asking for help is exactly the right move. Let’s break this down slowly, starting with Fermat’s and Rolle’s Theorems, then walk through your homework problems step by step.

Let’s start with plain-language explanations to cut through the confusion:

• Fermat’s Theorem: If a function $f(x)$ has a local maximum or minimum at some point $c$, and $f$ is differentiable at $c$, then $f'(c) = 0$. Think of it this way: if you’re at the peak of a smooth hill or the bottom of a valley, the tangent line there has to be flat (slope = 0).
• Rolle’s Theorem: This is a specific, practical extension of Fermat’s Theorem with three key conditions:
  1. $f(x)$ is continuous on the closed interval $[a, b]$
  2. $f(x)$ is differentiable on the open interval $(a, b)$
  3. $f(a) = f(b)$
     If all three are true, then there’s at least one point $c$ in $(a, b)$ where $f'(c) = 0$. In simple terms: if a smooth, unbroken curve starts and ends at the same height, it must level off somewhere in between.

Now let’s use these ideas to solve your two problems.

Problem 1: Prove $y = x^3 - 6x^2 + 12x - 8$ has at most two zeros #

First, find the derivative to see where the function’s slope is zero:
$$y' = 3x^2 - 12x + 12 = 3(x^2 - 4x + 4) = 3(x-2)^2$$
Notice $y' = 0$ only when $x=2$—this is the single point where the tangent line is flat.

We’ll use proof by contradiction here:

Suppose the function has three or more distinct zeros, say $a < b < c$, where $y(a) = y(b) = y(c) = 0$.

By Rolle’s Theorem, since $y(a)=y(b)=0$, there must be some $c_1$ in $(a,b)$ where $y'(c_1)=0$. Similarly, since $y(b)=y(c)=0$, there must be some $c_2$ in $(b,c)$ where $y'(c_2)=0$.

But we already found that $y'$ only equals zero at $x=2$—we can’t have two distinct points where the derivative is zero. This contradicts our initial assumption.

Therefore, the function can have at most two zeros. (As a side note, this function factors to $(x-2)^3$, so it actually has a single triple zero at $x=2$—but the theorem-based proof still holds!)

Problem 2: Prove $f(x) = x^5 + 5x^3 + 45x + 9$ has exactly one zero #

We’ll split this into two parts: proving there’s at least one zero, then proving there’s at most one zero.

Step 1: At least one zero (using the Intermediate Value Theorem) #

Polynomials are continuous everywhere, so the Intermediate Value Theorem applies:

• When $x$ approaches $-\infty$, the $x^5$ term dominates, so $f(x) \to -\infty$.
• When $x=0$, $f(0) = 0 + 0 + 0 + 9 = 9 > 0$.

Since $f(x)$ goes from negative infinity to a positive number, it must cross the x-axis at least once. So there’s at least one zero.

Step 2: At most one zero (using Rolle’s Theorem / monotonicity) #

First, find the derivative:
$$f'(x) = 5x^4 + 15x^2 + 45$$
Notice every term here is non-negative:

• $5x^4 \geq 0$ for all $x$
• $15x^2 \geq 0$ for all $x$
• $45 > 0$ always

This means $f'(x) = 5x^4 + 15x^2 + 45 > 0$ for every real number $x$. A function with a positive derivative everywhere is strictly increasing—it never decreases, so it can only cross the x-axis once.

To formalize this with Rolle’s Theorem, use contradiction again:

Suppose there are two distinct zeros $a < b$, where $f(a) = f(b) = 0$.

By Rolle’s Theorem, there must be some $c$ in $(a,b)$ where $f'(c) = 0$. But we just showed $f'(x)$ is always positive—this is impossible.

Combining both steps: the function has exactly one zero.

Hope this makes things click! If you’re still stuck on any part of the theorems or need help with similar problems, don’t hesitate to ask.

内容的提问来源于stack exchange，提问作者Zachary Mueller