When is the function not differentiable? (see graph attached)

The answer key says b, d, g ,h.

I don't understand why the function is not differentiable at g. I thought that functions can be differentiated at removable discontinuities.

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- Nov 26th 2009, 05:28 PMharmonicadudeecalc problem
When is the function not differentiable? (see graph attached)

The answer key says b, d, g ,h.

I don't understand why the function is not differentiable at g. I thought that functions can be differentiated at removable discontinuities. - Nov 26th 2009, 05:58 PMskeeter
- Nov 26th 2009, 05:59 PMtonio

It's aweird question: at b it is not diff. because, apparently, there's a vertical line there...but then that's not a function!

At d it is clear: the function has a "shpitz" or saw-tooth point there, so fine. At g and h the function isn't defined so clearly it can't be continuous there, no matter g is a removable discontinuity...

Tonio - Nov 26th 2009, 06:01 PMredsoxfan325
Derivatives can't exist where a function isn't defined. How would you draw a tangent line to a point that doesn't exist?

More formally, think about how you calculate the derivative at the point $\displaystyle g$.

$\displaystyle \lim_{h\to0}\frac{f(g+h)-f(g)}{h}$

But if $\displaystyle f(g)$ doesn't exist, you can't calculate that limit.

However, if you wanted to calculate $\displaystyle \lim_{x\to g}f'(x)$, that would exist (in this case). - Nov 26th 2009, 06:03 PMredsoxfan325
- Nov 26th 2009, 06:14 PMharmonicadudee
Well I guess my calculus teacher was wrong....

Thanks guys. - Nov 26th 2009, 06:44 PMtonio

What EXACTLY "isn't necessarily true"? $\displaystyle \sqrt[3]{x}$ is NOT a line segment at the origin. Since this is a visual exercise I am basing my opinion on what I see and that drawing shows like a vertical line segment at b. Of course, one could bring examples even more basic, as $\displaystyle \sqrt{x}$ , which also isn't differentiable at the origin.

Tonio - Nov 27th 2009, 09:08 AMredsoxfan325
Ok. We saw different things.