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I refer to a previous question: Directional Derivative Existence and the answer given.

I like his answer, however, I don't understand why we can use the Mean Value Theorem in that scenario.

Basically my concern is that it is only given that $f_x$ is continuous at a single point $(p,q)$, and $f_x$ exists at $(p,q)$. For the Mean Value Theorem, I believe we need continuity on a closed interval and differentiability on an open interval?

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yoyostein
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The Mean Value Theorem needs only that $f$ be continuous on a closed interval $[a,b]$ and that the limit in the definition of the (directional) derivative of $f$ exists in $(a,b)$, either as a finite real number or as $\pm \infty$. Notice that when the limit is finite, that limit is the derivative.

In particular, since you say $f_x$ is continuous at $(p,q)$, it is implied that $f_x$ (and hence the aforementioned limit) exists in a neighborhood of $(p,q)$. Hence, if $f$ is continuous near $(p,q)$ everything should be good.

Can you see that $f$ is continuous near $(p,q)$?

Fimpellizzeri
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  • How do I see $f$ is continuous near $(p,q)$? I know that differentiability implies continuity for single variable case. However I am having difficulty seeing it here. – yoyostein May 19 '16 at 05:44
  • Since you're applying the MVT only with respect to the $x$ variable, you need only continuity with respect to it. – Fimpellizzeri May 19 '16 at 13:14