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Rizags
 one year ago
Help with differentiation Please! Problem Below
Rizags
 one year ago
Help with differentiation Please! Problem Below

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Rizags
 one year ago
Best ResponseYou've already chosen the best response.0If\[f(x) = x^2, x \le1\] and \[f(x)=2x, x>1\] is f differentiable at 1?

Rizags
 one year ago
Best ResponseYou've already chosen the best response.0wait though, it isn't continuous at 1

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0your write cuz it looks like this dw:1444166725638:dw IT is continuous but not differentable at x=1 sorry if i confused u before

Rizags
 one year ago
Best ResponseYou've already chosen the best response.0also why is it not differentiable

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0Limits: 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)

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444167201542:dw

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0So when we approach the discontinuity form above and below point x=1 the graph no longer approaches the same value

Rizags
 one year ago
Best ResponseYou've already chosen the best response.0is this valid:\[f'(1)=\lim_{h \rightarrow 0^{}}\frac{ (1+h)^2(1)^2 }{ h }=2\] and\[f'(1)=\lim_{h \rightarrow 0^+}\frac{2(1+h)+2(1)}{h}=2\]

Rizags
 one year ago
Best ResponseYou've already chosen the best response.0because the limits are the same, its just that the function isn't continuous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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.

Rizags
 one year ago
Best ResponseYou've already chosen the best response.0that isn't the definition of continuity

Rizags
 one year ago
Best ResponseYou've already chosen the best response.0limits from both sides have to be the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a 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
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