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?

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 $$(0,3)$$.

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 $$(0,3)$$?

• I think your calculation in Ex2 is wrong Nov 10, 2019 at 8:09

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Nov 10, 2019 at 12:00

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

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.

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.