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Consider $f:\mathbb{R^2}\to\mathbb{R}$ defined by $$f (x,t)=\cases{\frac{\mathrm{sin}(xt)}{t} & $\text{ for } t \neq 0$\\ x& $\text{ for}\ t =0.$}$$ Use the Mean Value Theorm to show that $f(x,t) < x$ for all positve $t$ and $x$.

I know how to apply MVT to single variable functions but we have never been taught how to apply MVT to multi-variable functions like above so I really don't have any idea how to do this question.

Right now, my guess is to just let $x$ be some constant $x_0$ which is bigger than $0$, so that $f(x,t)$ just becomes a single variable function, $f(t)$. Then apply MVT on that. However not sure if that is correct or not.

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This looks like a parameterized family of single variable functions. I am supposing they want you to treat $f(x,t)$ as a function in $x$ with $t$ constant. Then the derivative with respect to $x$ (viewing $t$ as constant) is $\cos(xt)$.

If you apply the MVT to the interval from $0$ to $x$ you get

$$ \cos(x_0t)(x - 0) = \frac{\sin(xt)}{t} - \frac{\sin(0)}{t}. $$

Or to simplify,

$$ \cos(x_0t)x = f(x,t). $$

You can probably figure out the rest, just keep in mind that you are looking for when $\cos(x_0t) \ne 1$. I.e. you can't have $x_0t \in 2\pi \mathbf{Z}$. If $x$ is large enough that $x_0t = 2\pi$ is possible you will need to use another argument.

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  • $\begingroup$ Thank you for your answer. Just a couple of questions. Why did you decide to let 'x' be the variable and 't' the constant, was it just because it gives a nicer derivative? Also, is this method (of letting a variable be a constant) actually valid? Can you use this method for any continuous and differentiable two variable function? $\endgroup$ – Nanoputian Jun 1 '17 at 6:39
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    $\begingroup$ @Nanoputian There are two possibilities so you try both. In this case I used $x$ as the variable because , in my experience, it is more common for $x$ to be a variable than a constant. You can use this for two variable functions that are partially differentiable in whichever variable you choose to vary. You can also apply the MVT along lines in which a directional derivative exists. $\endgroup$ – Trevor Gunn Jun 1 '17 at 13:06

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