# Thread: Help Explain This (Mean Value Theorem)

1. ## Help Explain This (Mean Value Theorem)

I know that the mean value theorem applies when an interval [a,b] is continous and (a,b) is differentiable. When it does, there is a f(c) where f'(c) = f(b) -f(a) divided by (b-a)

I have the pic of the problem below for problem 1a and 1b

How am i suppose to find minimum possible value of f(2) if f is differentiable with f(0) = 1 and f'(x) >= -4 for -9 <= <=9?

I'm not sure what to do since the slope if variable. is it possible to calculate f(x) when slope is -4? I'm not sure what to do and dont understand what they are asking.

2. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by lc99
I know that the mean value theorem applies when an interval [a,b] is continous and (a,b) is differentiable. When it does, there is a f(c) where f'(c) = f(b) -f(a) divided by (b-a)
I have the pic of the problem below for problem 1a and 1b

How am i suppose to find minimum possible value of f(2) if f is differentiable with f(0) = 1 and f'(x) >= -4 for -9 <= <=9?
\begin{align*}[0,2]&\subset [-9,9] \\\exists c &\in[0,2]\\\dfrac{f(2)-f(0)}{2-0}&=f'(c)\ge -4\\f(2)-1&\ge-4\\f(2)&\ge-3 \end{align*}

3. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by Plato
\begin{align*}[0,2]&\subset [-9,9] \\\exists c &\in[0,2]\\\dfrac{f(2)-f(0)}{2-0}&=f'(c)\ge -4\\f(2)-1&\ge-4\\f(2)&\ge-3 \end{align*}
Post Script correction.
\begin{align*}[0,2]&\subset [-9,9] \\\exists c &\in[0,2]\\\dfrac{f(2)-f(0)}{2-0}&=f'(c)\ge -4\\f(2)-1&\ge-8\\f(2)&\ge-7 \end{align*}

4. ## Re: Help Explain This (Mean Value Theorem)

I''m kinda confused how the mean value theorem was applied? How does it help us find the minimum value cause i thought the mean value theorem helped us find the point on the graph where the the slope of that point is equal to the secant line at [0,2]

How do i know that -7 is the smallest value ? Is there a function that i can test it with so that it's actually the smallest?

5. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by lc99
I''m kinda confused how the mean value theorem was applied? How does it help us find the minimum value cause i thought the mean value theorem helped us find the point on the graph where the the slope of that point is equal to the secant line at [0,2]
How do i know that -7 is the smallest value ? Is there a function that i can test it with so that it's actually the smallest?
Do you even know what the mean value theorem says?
If you do, then tell where that theorem is used in the following.
\begin{align*}[0,2]&\subset [-9,9] \\\exists c &\in[0,2]\\\dfrac{f(2)-f(0)}{2-0}&=f'(c)\ge -4\\f(2)-1&\ge-8\\f(2)&\ge-7 \end{align*}~?~?~?.

The statement of the question tells us about $f'$ on $[-9,9]$. What is that property?
If you know that property, where was it used in the above?

6. ## Re: Help Explain This (Mean Value Theorem)

I learned that the mean value theorem applies when the function is continous and differentiable in interval [a,b] then there is f'(c) where it equals slope at points f(a) and f(x) . I kinda understand what you did there which was to find the value of f(2), but i dont really understand why keep the derivative to be -4 when we multiply? How do we make sure that this isn't any other number greater than -4? Do i just test numbers greater than -4?

I just talked to my teacher and he kinda explained to me how i had to find a function to prove that -7 is actually the minimum and that every other smaller value doesnt work.

7. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by lc99
I learned that the mean value theorem applies when the function is continous and differentiable in interval [a,b] then there is f'(c) where it equals slope at points f(a) and f(x) . I kinda understand what you did there which was to find the value of f(2), but i dont really understand why keep the derivative to be -4 when we multiply? How do we make sure that this isn't any other number greater than -4? Do i just test numbers greater than -4?
You are given that $f'$ that exists on $[-9,9]$ & moreover, $\forall x\in[-9,9],~f'(x)\ge -4$.
That tells us that $f$ meets all requirements for use of the mean value theorem on $[-9,9]$.

So on the interval $[0,2]$ we have
\begin{align*}\exists c &\in[0,2]\\\dfrac{f(2)-f(0)}{2-0}&=f'(c)\ge -4\\f(2)-1&\ge-8\\f(2)&\ge-7 \end{align*}
That clearly says that $f(2)\ge -7$. If you do not see that, you must say why.

8. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by lc99
I just talked to my teacher and he kinda explained to me how i had to find a function to prove that -7 is actually the minimum and that every other smaller value doesnt work.
According to the time-stamp, you added the above after I replied.
All I can say is that your teacher is simply wrong. But you have seen that before; look at your post on the value of $a$ so $fg$ is continuous. Were I you, I would complain about this teacher.

9. ## Re: Help Explain This (Mean Value Theorem)

Yeah. I'm confused on that. Sadly his office hours are cancelled and I can't ask. But, I forgot to mention that he said that I needed a function to prove the minimum value in order to fully complete my argument. I'm not sure what he means? I think I'm suppose to use the point at (2) to make a equation of a line.

10. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by lc99
Yeah. I'm confused on that. Sadly his office hours are cancelled and I can't ask. But, I forgot to mention that he said that I needed a function to prove the minimum value in order to fully complete my argument. I'm not sure what he means? I think I'm suppose to use the point at (2) to make a equation of a line.
That does not make sense. You cannot prove this type of statement by example. You can disprove it by counterexample, except there will not be one. Additionally, he needs to be very careful when stating that "a smaller number would not work". He is using the concepts of minimum value and greatest lower bound yet calling them both minimum values. He never says that the function achieves a derivative of -4 at any point. Anyway, I digress. If he really wants a function, I recommend the line with slope -4 that passes through the point (0,1).

11. ## Re: Help Explain This (Mean Value Theorem)

Yeah. He also mentioned that if we find a contradiction, then we argue or something. Idk. Something about contradiction in this problem

This is the email he sent us in regards to this problem :

"Do problem 1 from today's workshop. Make sure that your argument for part a is complete. (To prove that some value is minimal, you should come up with a function which actually shows that this value is possible; and that every smaller value doesn't work)."

12. ## Re: Help Explain This (Mean Value Theorem)

Originally Posted by lc99
Yeah. He also mentioned that if we find a contradiction, then we argue or something. Idk. Something about contradiction in this problem

This is the email he sent us in regards to this problem :

"Do problem 1 from today's workshop. Make sure that your argument for part a is complete. (To prove that some value is minimal, you should come up with a function which actually shows that this value is possible; and that every smaller value doesn't work)."
That is very different from what you said above. What he is saying is demonstrate that it is possible to achieve the minimal value you calculate. The line I gave you demonstrates this very well. Since the line I gave has the property that $f'(x) = -4$ for all $x$, it is not possible for the function to reach a smaller value. This line uses the minimum possible derivative for all $x$. The derivative represents the change in y as x changes. You know the value of $y$ when $x=0$. So, if the minimum derivative is -4, then as x increases by 2 (going from 0 to 2 is an increase of 2), y can change by a maximum of $2\cdot(-4) = -8$ (the decrease for $y$ of maximal magnitude is $-8$). That is what a derivative means.

Think of it like you are trying to determine the minimum stopping distance for a car. You know that the car's stopping power is based on the coefficient of kinetic friction the car makes with the ground. From this, you can determine the maximum deceleration the car may experience. If at time t=0, you know the car is going 50 mph, you want to find what is the minimum speed at various times after that point (knowing the maximal stopping power), you are basically looking at this exact same problem. This has real world applications.

13. ## Re: Help Explain This (Mean Value Theorem)

Hey,
I was wondering if i am explaining my answers for part b right?

or can i just explain that: f'(x) can be larger than -4, so if the f'(x)>0, f(5) will always be positive.

for f(5) = -20, i solved for f'(c) = (f(5) - 1)/5 = -4.2, so it is not possible for f(5)=-20 when slope has to be -4.2 in the equation y=(19/5)x + 1 since it doesn't satisfies that f'(x) >= -4 and f(0) = 1.

for f(5) = -16, i solved for f'(c) = (f(5) -1)/5 = -17/5 this time plugging f(5) = -16. Therefore, f'(c) should be -17/5 which is possible because it is greater than -4.

(I was wondering if i fully explained the problem? , or am i missing something important?)
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}