I'm lost on how to solve this. Can someone explain please?
Suppose f(x)=2 for x<3 and f(x)= -4 for x>3
A) what is f(3) if f is left continuous at x=3
B) what is f(3) if f is right-continuous at x=3?
Suppose f(x)=2 for x<3 and f(x)= -4 for x>3
A) what is f(3) if f is left continuous at x=3
B) what is f(3) if f is right-continuous at x=3?
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So initially at f(3) we have a discontinuity which is why we have a piece wise function.
A) As we approach f(3) from the left, that means we are approaching from the negative side which means its x<3...
So A is 2
B) the exact opposite: -4
A) As we approach f(3) from the left, that means we are approaching from the negative side which means its x<3...
So A is 2
B) the exact opposite: -4
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a) 2
b) -4
i've never heard of it put this way but it sounds like left continuous from a graphical perspective means it retains the value approaching from the lower value just before the jump discontinuity and visa versa for the right.
b) -4
i've never heard of it put this way but it sounds like left continuous from a graphical perspective means it retains the value approaching from the lower value just before the jump discontinuity and visa versa for the right.