1. calc 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.

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.
you thought wrong. if a function is differentiable at a point, then it is also continuous at that point.

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.

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

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.
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 $g$.

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

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

However, if you wanted to calculate $\lim_{x\to g}f'(x)$, that would exist (in this case).

5. Originally Posted by tonio
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
That's not necessarily true. What about $\sqrt[3]{x}$ at the origin?

6. Well I guess my calculus teacher was wrong....

Thanks guys.

7. Originally Posted by redsoxfan325
That's not necessarily true. What about $\sqrt[3]{x}$ at the origin?

What EXACTLY "isn't necessarily true"? $\sqrt[3]{x}$ 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 $\sqrt{x}$ , which also isn't differentiable at the origin.

Tonio

8. Ok. We saw different things.