# Does MVT prove the existence of f'(c)=0

Because MVT states that : $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

This proves that the derivative is EQUAL to the secant line through $a$ and $b$.

Is it possible for $f'(c)=0$ if $f(b) \neq f(a)$, given that $f(c)< f(a)$ or $f(c) > f(a)$ for a single $c$ where $b>c>a$.

EDIT: I should be more concrete. I have been given a problem with only points of data.
The table is as follows: $$f(0)=20 \\ f(5)=17,\\ f(10)= 10,\\ f(20)=10, \\f(25)=16, \\f(30)=21$$

This is similar to an AP Calculus BC FRQ problem; I just changed the numbers. How many instances can $f'(x)=0$.

I know Rolle's theorem states that $f'(c)=0$ between $x=10$ and $x=20$.

But what I care about is if MVT proves that there could be $f'(c)=0$ when $f(0) \neq f(30)$

• I can't really understand what you are asking, but it seems like you might be interested in Rolle's theorem – user139388 May 3 '14 at 7:01
• Are you thinking about Rolle's theorem? – IAmNoOne May 3 '14 at 7:27

I see your confusion, consider $f(x)=1$ and you will get $f'(c)=0$.

You wouldn't consider the case $a=b$ you would consider the limit $b\to a$ which would give you the derivative at $a$.

Also it does not prove that the derivative is always tangential to the secant line, it just proves that there is at atleast one point, namely $c$ in $[a,b]$, where they are tangential.

But to bring things into context with your data table, if you only consider the end points $a=0,b=30$, then MVT Tells us:

$\exists c\in(0,30):f'(c)=\frac{f(10)-f(0)}{30-0}=\frac{21-20}{30}=\frac{1}{30}$

But if we restrict to $[10,20]$ then MVT tells us:

$\exists c\in(10,20):f'(c)=\frac{f(20)-f(10)}{20-10}=\frac{10-10}{10}=0$

But $[10,20]\subset[0,30]$ so in this sense MVT shows that $\exists c\in(0,30):f'(c)=0$

Extra part: assume we only had: $f(0)=20,f(5)=17,f(25)=16,f(30)=21$ and assume $f$ is continuously differentiable on the whole interval $(0,30)$

MVT tells us, $\exists d\in[0,30]:f'(d)=\frac{f(30)-f(0)}{30-0}=\frac{1}{30}$

And MVT tells us $\exists d^*\in[5,25]:f'(d^*)=\frac{f(25)-f(5)}{25-5}=-\frac{1}{20}$

Now Intermediate value theorem tells us that $\exists c^*\in[0,30]:f'(c^*)=0$

(due to the change in sign of the derivative, that we have assume to be continuous)

So here the combination of MVT and IVT gives you your desired result, but it really does depend on the given data set, since if we only had the two end points $(0,30)$ all we would gain from MVT is that there is a point $c\in[0,30]:f'(c)=\frac{1}{30}$

• I dont understand our first two sentences. "Also it does not prove that the derivative is equal to the secant line" I dont know what you're saying about this. tutorial.math.lamar.edu/Classes/CalcI/MeanValueTheorem_files/… This picture shows that $f'(c) = \frac{f(b)-f(a)}{b-a}$ – Guestx1000020 May 3 '14 at 8:46
• I meant the derivative is equal to the slope of (parallel to)the secant line. Also, can you show an image of what you mean with "it just proves that the two intersect at atleast one point, namely c in [a,b]" – Guestx1000020 May 3 '14 at 8:53
• In that sentence, I'm saying that mean value theorem does not imply that the derivative equal to the slope secant line on the whole interval $[a,b]$, since if $f$ is not linear for instance if it is a cubic, then its gradient is not constant. But there is a point in the interval where the derivative is equivalent to the gradient of the slope, i.e. there is at least one point where they are tangential – Ellya May 3 '14 at 8:56
• I actually do understand your first sentence but not your second. Also, given that I gave a more concrete question, I kindly ask if you could assist with the new question. – Guestx1000020 May 3 '14 at 8:59
• @Guestx1000020 Yes I have put in an edit, that handles your data table :) – Ellya May 3 '14 at 9:02