Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4).
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Originally Posted by spikedpunch Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4). Notice that . If then .
could you explain why delta is min( 1, epsilon/5 )?
Originally Posted by spikedpunch could you explain why delta is min( 1, epsilon/5 )? Well you want it to work.
So I can see this works if if the min is epsilon/5, but how would it work if the min is 1? That is, how can I show that the f(x)-f(t)<epsilon?
Originally Posted by spikedpunch [COLOR=#0000AA][FONT=monospace] So I can see this works if if the min is epsilon/5, but how would it work if the min is 1? Just think about it? If then , right?
I guess I am confused as to why you showed -1<x-2<1 in the first place. If 1 is the min, then wouldn't it be absv(x^2-4)=absv(x+2)absv(x-2)=(1)(5)=5...which is not epsilon?
Originally Posted by spikedpunch I guess I am confused as to why you showed -1<x-2<1 in the first place. If 1 is the min, then wouldn't it be absv(x^2-4)=absv(x+2)absv(x-2)=(1)(5)=5...which is not epsilon? I think that you need to do some background work on this.
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