Mean value theorem contradiction

Question

Let $f(x) = 2 − |2x − 1|$. Show that there is no value of $c$ such that $$f(3) − f(0) = f'(c)(3 − 0).$$ Why does this not contradict the Mean Value Theorem?

Note: I am not asking how to solve this problem, I'm asking a question about this problem.

EX1: For the answer of this question, $x$ is not differentiable at $1/2$ because the left slope $2$ is not equal to the right slope $-2$. So there is no value $c$ where $f'(c)$ has a slope of $-4/3$ on the interval $(0,3)$.

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EX2: My question is that what if you restricted the domain to $[3,9]$? Then the function would be continuous as before but also DIFFERENTIABLE this time. However, there is still no value $c$ where $f'(c)$ has a slope of $-4/3$.

I know my question may sound confusing but in short, what does it really mean when it says "contradict the mean value theorem?" Is EX2 a contradiction to MVT and EX1 isn't because EX1 has an extra condition where it's not differentiable on the domain $(0,3)$?

Answer 1

You may not apply Mean Value Theorem to $f(x)= 2 − |2x − 1|$ with respect to any interval which does contain the point $1/2$ where $f$ is not differentiable (and therefore the hypotheses of the formal statement are not satisfied). If the interval is $[0,3]$ then we have a contradiction because there does not exist $t\in (0,3)$ such that $$-4/3=\frac{f(3)-f(0)}{3-0}=f'(t).$$

On the other hand you may apply Mean Value Theorem to $f$ with respect to any interval which does not contain the point $1/2$.

Since $1/2$ does not belong to $[3,9]$, by MVT there is $t\in (3,9)$ such that $$\frac{f(9)-f(3)}{9-3}=f'(t)$$ which holds because $(-15-(-3))/(9-3)=-2$. Actually, in this case, $\frac{f(9)-f(3)}{9-3}=f'(t)$ holds for ANY $t\in (3,9)$ because $f$ restricted to the interval $[3,9]$ is the line $y=2-(2x-1)=-2x+3$ and $f'$ is identically constant with value $-2$ on that interval.

Answer 2

$$f(x) = 2 - |2x-1|$$

For $x \le \dfrac 12$, $f(x)=2-(-2x+1)=2x+1$ and $f'(x)=2$..

For $x \ge \dfrac 12$ , $f(x)=2-(2x-1)=-2x+3$ and $f'(x)=-2$.

So $f(x)$ is not differentiable at $x=\dfrac 12$.

Robert Z has answered the second part.

Answer 3

If you restrict the domain of $$f(x) = 2-|2x-1|$$ to $[3,9]$ then your function is simply $$f(x) = 2-(2x-1) = 3-2x $$ which is a straight line with slope of $-2$

The linear function satisfies mean value theorem at every point of the interval that is $$\frac {f(b)-f(a)}{b-a}= \frac {3-2b-3+2a}{b-a} =-2$$ which is your derivative at every point of the interval.