Question

How would I approach this problem? (Mean Value Theorem with a system of equations, no specific closed set given but i'm assuming it's implied through the set given for each equation.)

*Below*

Answer 1

\[g(x) = x ^{3}, -2\le x \le 0 --- g(x) = x ^{2}, 0 < x \le 2\]\[g(x) = 3x ^{2}, g(x) = 2x\]Now how do I apply two derivatives to the Mean Value Theorem equation???\[f'(c) = \frac{ f(b) - f(a) }{ b - a }\]Do I just apply the ends of the ranges of each? And then, even if I did that, how do I apply that with less than/less than or equal to signs? I don't have a closed set.

Answer 2

WHOOPS, the second line are the derivatives for each one, respectively. Forgot the aopstrophe

Answer 3

what is the domain its asking for the MVT on?

Answer 4

It isn't. That's why i'm super confused, lol.

Answer 5

I'll take a snapshot to make sure i'm not crazy. One minute.

Answer 6

whats the original question? im lost too haha.

Answer 7

(One second, learning how to use adobe snapshot -_-

Answer 8

haha ;P

Answer 9

Jesus christ, i'm totally stumbling over myself like a pro at the moment, all the other ones give a specific domain.

Answer 10

wonderful title for the document

Answer 11

ok i get it.

Answer 12

the domain is from -2 to 2

Answer 13

im assuming you add the derivatives together.

Answer 14

if you graph em you get a line for the derivatives.

Answer 15

I'm gonna wait until somebody who's super, super sure about the adding derivatives (or just what to do with them) comes around. This isn't of any consequence (it's a practice problem, it's not due, but there's no answer given) but i'd nonetheless like to know for sure. Thanks, though!

Answer 16

haha, its been awhile since ive derived a piece-wised defined

Answer 17

If you are talking about number 8 (which I believe I am following this correctly) then @yoyointegrals is correct in that you basically just add the domains together. (assuming they are continuous, which if you look at a graph of x-squared and x-cubed, they would be since the "shared" point is at 0)

Answer 18

the derivatives are continuous.

Answer 19

Add the domains, or add the functions? Because although it's mainly just petty semantics, if I take that literally, I think of adding plus or minus infinity to plus or minus infinity, which makes basically no sense at all. Who knows, my semantics are in all likelihood worse than yours, but that doesn't sound consistent with whatever i've been taught.

Answer 20

Haha. no, in the piecewise function, since both parts of it are continuous you can look at it from the interval [-2, 0]U(0, 2] that make sense?

Answer 21

b = 2, a = -2

Answer 22

since both parts of it are continuous with each other* you can use it for the MVT

Answer 23

define f(b) and f(a) within their respective domains.

Answer 24

Yeah, I thought about that, I just don't understand what to do with the derivatives. I had always suspected that the interval should just be considered [-2, 2], but I still don't quite understand to do with the derivatives...?

Answer 25

hmm, I'd need to review the MVT to be able to say for sure. But the domain is definitely [-2, 2]

Answer 26

Yeah. Well thanks for the help for the time, guys. This is a pretty weird instance of it, so, I dunno, hopefully some mathematician god on the forum or something will drop in.

Answer 27

Hmm, looking over the MVT, it shouldn't particularly matter that you were given "two" functions to define g(x), but you'd just need to be careful when applying the theorem that you use the correct one for each domain.

Answer 28

So, we can see the MVT on the paper, no sense restating it.
f(b) would then be \[2^3\] and f(a) would be \[-2^2\]since you'd have to evaluate each value for the function defining it's respective domain. Does that make sense?

Answer 29

Yup.

Answer 30

We can solve the right hand part\[\frac{ f(b) - f(a) }{ b-a }\] fairly easily then, it's just the deal with the derivatives. And I don't get precisely what you mean when you say being "careful" with regards to applying the proper range to each one. I have a vague idea of it, but no idea what it means in practice.

Answer 31

Well, basically the function on the interval [-2, 0] is defined as x-squared, so if we find the value for a number on the interval we have to use x-squared and not x-cubed. (sorry about the delay, got side tracked.)

Answer 32

And what do you mean the deal with the derivatives?

Answer 33

Ugh, now i'm confused. What i'm thinking you're saying is that for a and b, for both derivatives, we should use [-2, 2]. And then take the endpoints applicable to each derivative's original function's domain and plug them in? I feel like you're saying this for the first oen for example:\[f(x) = x ^{3}, f'(x) = 3x ^{2}, 3x ^{2} = \frac{ f(b) - f(a) }{ b - a }\] Where b and a are 2 and -2, respectively. But that seems horribly wrong (I think i'm just misinterpreting you), I feel like it should be the exact same thing where b and a are 0 and 02, respectively.

Answer 34

The way i'm thinking about it now, it would be permissible to have-OH, "find the value or values", so yes, more than one is definitely permitted, but not maybe in the same sense; the way i'm thinking about it now, each piecewise function could have its own closed interval, but that seems to totally disagree with the very idea of the Mean Value Theorem.

Answer 35

*0 and 2, respectively.

Answer 36

I would imagine that since you are looking for the c values, that you would basically treat it as two separate problems essentially. Where part 1 would use g(x) = x-squared on the interval [-2, 0] and part 2 would use g(x) = x-cubed on the interval (0, 2].

Thought I would agree that that still doesn't seem quite satisfactory...

Answer 37

Yeah, I agree on that too, it just seems strange, uh, let me think about this. Well, jeez, I guess they are continuous technically, but that's, that's just...it doesn't seem right, lol. Anyways, thanks. I'll ask my professor next class.

Answer 38

Wait, sorry, I might want to hold off on just shutting the book, lol. Somebody might show up over the weekend who knows the answer, so i'll keep the question open for now.

Answer 39

Yeah, I'd probably do that. if it was just solve for f '(c) then it would be fairly easy. But the exact value(s) for c I imagine you would still want the full interval... I might come back to it if I have time over the weekend.

Answer 40

Np, thanks dude.

Answer 41

Ah! I think I've got it. It's fascinating. One sec.

Answer 42

So, what the MVT basically tells us is that there is some tangent line parallel to the secant line connecting points a and b. So then, what we can do it solve the MVT to find f '(c) (the slope of the tangent line. I'll call it m) Then, we know that there will be some c in which f '(c) equals m.

Answer 43

So, solving the MVT for f '(c) we get \[\frac{ f(b)-f(a) }{ b-a } = \frac{ g(2)-g(-2) }{ 2-(-2) }=\frac{ 2^2-(-2)^3 }{ 2-(-2)}=\frac{ 4-(-8) }{4} = \frac{12}{4} = 3\]

Answer 44

Then, to find the value(s) for c, we set the derivative of x-cubed and x-squared equal to 3 and solve. So \[g '(x) = 3x^2 [-2, 0]\] and \[g '(x) = 2x (0, 2]\] Solving each for 3 we find that on the [-2, 0] it =1 and on the interval (0, 2] it equals 1.5. Since 1 is not on the interval [-2, 0] c must = 1.5.

Answer 45

Graphing the piecewise function and drawing the secant line connecting point a to point b we see that the secant line is also the tangent line of the equation, and it appears to cross at 1.5. So, I am pretty confident in this answer. Hope you understood all of that!

Answer 46

Give me like, 30 minutes. My brain kind of just imploded halfway through your answer.

Answer 47

Haha. Take your time, and ask about whatever you aren't following.

Answer 48

@blarghhonk8 [Achievement Get]: Acquire Open Study Hero status. http://www.youtube.com/watch?v=c7tzi8wkYgI

Answer 49

(And yeah, I got it. And, if you thought that up by yourself, congratulations on actually having determination, and for the most part proving it, unlike me, lol.)

Answer 50

Haha. I used some references. http://tutorial.math.lamar.edu/ is a great math resource. That was where I got the idea to graph it, and from there I found that the tangent line and secant line would have to be the same. And then finally realized that it would just work by plugging in the slope to both derivatives like you would any other MVT question like this and check to make sure the domain works.