# Examples where Rolle's Theorem fails for following

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## Solution 1

Assuming $a<b$. For part (a) one example is $f(x)=|x-(a+b)/2|.$ For (b) note that if $f$ is differentiable on $(a,b)$ then it is continuous on $(a,b)$ but not necessarily at $a$ or at $b.$ For example $f(x)=x$ for $x\in (a.b),$ and $f(a)=f(b).$ Regardless of the value of $f(a),$ this function $f$ can't be continuous at both $a$ and $b.$ For (c) one example is $f(x)=x$ .

## Solution 2

For (b), differentiability implies continuity only at points where the function is differentiable. In this case we are given $f$ is differentiable in $(a,b)$, so it can still be discontinuous at $a$ or $b$.

## Solution 3

Note that if $f$ is not continuous at $a$ or $b$, then it is possible that $f(a)$ or $f(b)$ is undefined. Then Rolles Theorem cannot be used.

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### David

I like video games and MountDew. I am studying to be a secondary math teacher and I love computers. Pure Math (BA).

Updated on August 01, 2022

### Comments

• David 3 months

Rolle's Theorem: Suppose that $f$ is continuous on $[a,b]$ and is differentiable on $(a,b)$. If $f(a) = f(b)$, then there is a number $c \in (a,b)$ for which $f'(c) = 0.$

Give an example where Rolle's Theorem fails if the following hypothesis is omitted:

(a) $f$ is continuous on $[a,b]$ and $f(a) = f(b)$ but $f$ not differentiable on $(a,b)$.

Example: $f(x) = |x|$

(b) f is differentiable on $(a,b)$ and $f(a) = f(b)$, but $f$ not continuous on $[a,b]$

Example: differentiability implies contininous not sure.

(c)$f$ is continouous on $[a,b]$ and differentiable on $(a,b)$ but $f(a) \neq f(b)$

Example: $f(x) = \sin{(x)}$

• Seven almost 7 years
It is given that $f(a) = f(b)$, so $f$ has to be defined at both $a$ and $b$.