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Define the function $f\colon \Bbb R^3\to \Bbb R$ by $$f(x,y,z)=xyz+x^2+y^2$$ The Mean Value Theorem implies that there is a number $\theta$ with $0<\theta <1$ for which $$f(1,1,1)-f(0,0,0)=\frac{\partial f}{\partial x}(\theta, \theta, \theta)+\frac{\partial f}{\partial y}(\theta, \theta, \theta)+\frac{\partial f}{\partial z}(\theta, \theta, \theta)$$

This is last question. I dont have any idea. Sorry for not Writing any idea. How can we show MTV for that question?

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Can you post the formulation for the MVT that you're familiar with? – Git Gud Jul 11 '13 at 20:20
$f(x+h)-f(x)=<D(f(x+\alpha h),h>$ with a number $\alpha \in(0,1)$ @gitGud – B11b Jul 11 '13 at 20:23
@Rsm1 Check Peterson's hint. Has he meantions in the comments, you're basically just dealing with a one variable function on disguise because you're evaluating $f$ at points in which all coordinates are equal. – Git Gud Jul 11 '13 at 20:31
Well, how can I apply MVT? This part is so important. @GitGud – Mathlover4 Jul 11 '13 at 20:34
up vote 2 down vote accepted

Following up on Peterson's hint, forget about the MVT for several variables and focus on the one dimensional version of it.

Consider the function $\varphi\colon [0,1]\to \Bbb R, t\to t^3+2t^2$.

The MVT guarantees the existence of $\theta\in ]0,1[$ such that $\varphi '(\theta)=\varphi(1)-\varphi (0)$.

Now try to relate $\varphi (1)$ with $f(1,1,1)$, $\varphi(0)$ with $f(0,0,0)$ and $\varphi '(\theta)$ with $\displaystyle \frac{\partial f}{\partial x}(\theta, \theta, \theta)+\frac{\partial f}{\partial y}(\theta, \theta, \theta)+\frac{\partial f}{\partial z}(\theta, \theta, \theta)$.

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Now I understand. Thank you:)) – Mathlover4 Jul 11 '13 at 20:47
I expect more complicated solution b'cuz of MVT – Mathlover4 Jul 11 '13 at 20:48
@Rsm1 No problem. – Git Gud Jul 11 '13 at 20:51
Sorry @GitGud I have another question. How do we find the value $\theta$? Please help me:( – Mathlover4 Jul 12 '13 at 13:35
@Rsm1 Just use the fact that $0<\theta <1$ and $\varphi '(\theta)=\varphi(1)-\varphi (0)$. – Git Gud Jul 12 '13 at 16:13

Hint: Consider $g(t):=f(t,t,t)$. What is $g'(t)$?

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$g'(t)=3t^2+4t$ – B11b Jul 11 '13 at 20:25
Well, yes. I was more referring to noticing its format without using the specific definition of $f$: $$g'(t)=f_x(t,t,t)+f_y(t,t,t)+f_z(t,t,t),$$ by the chain rule. This function has one variable, so you can use the MVT on $g$ to establish the result on the page. (Note that the question asked isn't actually shown in your image!) – Nick Peterson Jul 11 '13 at 20:27
Please can you show me MVT application ? – Mathlover4 Jul 11 '13 at 20:32
The application of MVT is on $g$ and $t$ as @nrpeterson mentioned. Then "translate" that to a result in terms of $f$, $x$, $y$, and $z$ – angryavian Jul 11 '13 at 20:51

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