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I am trying to understand why the following question is correct. Regarding to the MVT.

Given: Find a number c satisfying the conclusion of the Mean Value Theorem

  1. $f(x)= x^{1/3}$, on $[-1,2]$
  2. $f(x)= x^{1/3}$, on $[0,2]$

For 1) I found that $f(x)$ is continuous and but not differentiable at $x=0$. For 2) I found that $f(x)$ is continuous and differentiable on $[0,2]$

$$ f'(x)= 1/3 \, x^{-2/3} $$

$$ \frac{f(b)-f(a)}{b-a} = \frac{f(2)-f(-1)}{2+1} = \frac{2^{1/3} + 1}{3} = 0.7533070166 $$ I found $C= 0.29$

For 2) I found $f(x)$ is continuous and differentiable on interval $[0,2]$.

$$ f'(x)= 1/3 \, x^{-2/3} $$

$$ \frac{f(b)-f(a)}{b-a}= \frac{f(2)-f(0)}{2-0}= 0.6299605249 $$ I found $c= 0.3849001795$

What i don't understand is the differentiability. Especially for 1) it's not differentiable at $x=0$ but for 2) it's differentiable.

The mean value theorem states that 1) continuous on $[a,b]$ 2) differntiable on $(a,b)$ and 3) for at least one value $c$ in $(a,b)$ s.t. $$ f'(c)=\frac{f(b)-f(a)}{b-a} . $$

For 1) function is continuous. there is at least one value $c$ in $[-1,2]$. Here is what I dont understand, function is NOT differentiable on $(-1,2)$. why? for 2) it's differentiable on $(0,2)$. my professor said that's bc it's not differentiable at $x=0$ for 1).

I am confused as to know how we can know when function is not differentiable at $x=0$? Do i just plug $x=0$ into the function? Do we graph first and then see if the graph displays cups or corner? like for 1) and 2) i was able to know that function is continuous and there is at least one value $C$ in $(a,b)$ s.t. $f'(c)= f(b)-f(a)=(b-a)$. I found the value of $C$ for both 1) and 2). I was able to take derivative of $f(x)$. for 1) it's not differentiable but for 2) it's differentiable. I thought it was differentiable for both since i was able to take derivative of the function..

help please..It bothers me that i do not understand clearly what she meant by not differentiable at $x=0$. ugh.

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  • $\begingroup$ Both functions are not differentiable in $x=0$ $\endgroup$ – Quickbeam2k1 Oct 4 '14 at 18:18
  • $\begingroup$ then 1) nd 2) fail to meet the MVT, right? but how come 2) is differentiable? but not differentiable for 1) $\endgroup$ – chris Oct 4 '14 at 18:21
  • $\begingroup$ Just for clearification: The MVT states that IF some conditions are fulfilled, then there is a point $c$ inside the domain such that the intermediate slope is attained. That is all. You still may find a point $c$ such that the intermediate slope is attained even though the functions is not differentiable (inside its domain). However, if there is no such point $c$ find such a point, then $f$ is not differentiable everywhere inside its domain. ($(A\Rightarrow B=\Longleftrightarrow (\lnot B \Rightarrow \lnot A)$ $\endgroup$ – Quickbeam2k1 Oct 4 '14 at 18:45
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The function $f(x)=x^{1/3}$ is not differentiable in $x=0$. However, the mean value theorem can be applied to your second case since $f$ is continuous on $[0,2]$ and differentiable on $(0,2)$. Check precisely the requirements of the MVT.

The differentiability on $(0,2)$ follows since the formula by Dr. Sonnhard Graubner in the other answer holds.

Differentiabily is defined pointwise. You have to check every point insinde the domain of the function if the difference quotients converge. This does not happen in $x=0$ as we infer from the following argument. If $f$ was differentiable in $x=0$ then the limit $$\lim_{h\to 0}\frac{f(h)-f(0)}{h-0}=\lim_{h\to 0}\frac{f(h)}{h}$$ must exists. Here $h$ is in $[-1,2]\setminus\{0\}$ or in $(0,2]$.

However, we find $$\frac{f(h)}{h}=\frac{1}{h^{2/3}}\xrightarrow{h\to 0}\infty.$$ Thus the difference quotient does not converge.

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  • $\begingroup$ thats my question. HOW? $\endgroup$ – chris Oct 4 '14 at 18:23
  • $\begingroup$ it seems like it has to do with the interval $\endgroup$ – chris Oct 4 '14 at 18:23
  • $\begingroup$ If the interval was let's say [-5,5] then f(x) is still not differentiable bc x=0 correct? $\endgroup$ – chris Oct 4 '14 at 18:25
  • $\begingroup$ what does bc mean? $\endgroup$ – Quickbeam2k1 Oct 4 '14 at 18:28
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    $\begingroup$ For the mean value to be applicable, the function in question only has to be differentiable on the interior of the interval on which it is defined (but continuous on the whole interval). Hence, $f$ is not differentiable in $0$ in both cases, but in the second it does not matter, because $0$ is not in the interior of $[0,2]$. $\endgroup$ – PhoemueX Oct 4 '14 at 18:32
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$f'(x)=\frac{1}{3x^{2/3}}$ is for $x=0$ not defined.

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    $\begingroup$ Even though your function is not defined in zero, this does not imply that $f$ is not differentiable at $x=0$. $\endgroup$ – Quickbeam2k1 Oct 4 '14 at 18:18
  • $\begingroup$ hello, it is the first derivative you can also consider $\frac{f(h)-f(0)}{h-0}=\frac{h^{1/3}}{h}=\frac{1}{h^{2/3}}$ $\endgroup$ – Dr. Sonnhard Graubner Oct 4 '14 at 18:26

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