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Use the Mean Value Theorem to prove that if a continuous and differentiable function $f$ has two roots, then $f'$ has at least one root.

My thought process so far:

The Mean Value Theorem tells us that if function $f$ is continuous and differentiable on an interval $[a,b]$, there is at least one $c\in [a,b]$ where

$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

If $f$ has two roots, it means that $f$ takes the value $0$ at two points.

If $f$ is continuous on an interval $[a,b]$ and has two roots, this means that there is a point $c\in [a,b]$ where $f'(x)=0$.

I'm thinking about approaching this task with a contradiction proof and assume the opposite: if $f(x)=0$ two times, then $f'$ never takes the value $0$ (I'm not sure if this is the best way though).

I'm getting stuck here and I'm not quite sure how the MVT is applicable to this task. At first, I was thinking writing out the form of the functions which do have two roots for example $ax^2+bx+c$ but doesn't the degree of the polynomial only tell you how many roots it has at least?

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    $\begingroup$ Suppose you had $f(a)=f(b)=0$. Does this help? $\endgroup$ Dec 11, 2017 at 8:31

3 Answers 3

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Let $f(x)$ have two roots at, say, $x_1$ and $x_2$. It thus follows that $f(x_1)=f(x_2)=0$.

Hence, by Rolle’s Theorem, we have that there is a number $c_1$ in $(x_1,x_2)$ such that $f’(c_1)=0$.

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It is possible to not use the Mean Value Theorem. Suppose $f'(x)$ has no roots - that is $f(x)$ is strictly increasing or strictly decreasing. Then by the intermediate value theorem, $f(x)$ only has one root.

Therefore, if $f(x)$ has two roots, then $f'(x)$ cannot have no roots, or in other words has at least one root.

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Given f is differential on R (1)

A differential function is necessarily continuous

=> f is continuous on R (2)

Let x1, x2 be roots of f that means f(x1) = f(x2) = 0 (3)

f satisfies all 3 hypotheses of Rolle's Theorem, we have a c on (x1, x2) such that f '(c)=0

Meaning that f ' have at least one root which is c

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  • $\begingroup$ Hi Cggg. Your notation R (1) and R (2) doesn't correspond to anything I see in the question. What does it mean? $\endgroup$
    – 311411
    Nov 2, 2021 at 0:15
  • $\begingroup$ The post is identifying with (1), (2), (3) the hypotheses of Rolle's Theorem. It's not R(1). It's just (1). $\endgroup$ Nov 2, 2021 at 0:31

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