## anonymous one year ago Help with differentiation Please! Problem Below

1. anonymous

If$f(x) = x^2, x \le1$ and $f(x)=2x, x>1$ is f differentiable at 1?

2. anonymous

wait though, it isn't continuous at 1

3. marigirl

your write cuz it looks like this |dw:1444166725638:dw| IT is continuous but not differentable at x=1 sorry if i confused u before

4. anonymous

that isnt continuous

5. anonymous

also why is it not differentiable

6. marigirl

Limits: usually we look at a specific value when the graph approaches a number that we are after. A limit won't exist when we get different solutions from above and below graph (See my picture)

7. marigirl

|dw:1444167201542:dw|

8. marigirl

So when we approach the discontinuity form above and below point x=1 the graph no longer approaches the same value

9. marigirl

same y value*

10. anonymous

is this valid:$f'(1)=\lim_{h \rightarrow 0^{-}}\frac{ (1+h)^2-(1)^2 }{ h }=2$ and$f'(1)=\lim_{h \rightarrow 0^+}\frac{2(1+h)+2(1)}{h}=2$

11. anonymous

because the limits are the same, its just that the function isn't continuous

12. marigirl

@surjithayer

13. anonymous

It is Continuous because the x value exists at all points. x=1 is subsidized for the leap. just because there is a leap doesn't make it not continuous.

14. anonymous

that isn't the definition of continuity

15. anonymous

limits from both sides have to be the same

16. anonymous

a function f is said to be continuous at x=a 1. Limit exists Means left hand limit =Right hand limit. 2. f(x)=f(a) Here limit does not exist at x=1 so it is not continuous at x=1 so it is not differentiable at x=1