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Functions, Limits, Continuity

Problems for "Continuity"

Continuity

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Problem : Is f (x) = x 3 + 2x + 1 continuous at x = 2 ?


Yes. All polynomial functions are continuous. To check,

f (x) = 13 and f (2) = 13    

Problem : Is f (x) = continuous at x = 4 ?


No. f (4) is undefined.

Problem : Is the following function continuous?

f (x) =    


Since this function is a junction of two continuous functions, we only have to worry about discontinuity at the point where the functions meet, i.e. at x = 2 .


f (x)   = x + 3 = 2 + 3 = 5  
f (x)   = (x 4 -11) = 24 - 11 = 5  


Thus, f (x) = 5 , and f (2) = 5 , so f (x) = f (2) . It follows that f is continuous.

Problem : Is the following function continuous?

f (x) =    


Since this function is a junction of two continuous functions, we only have to worry about discontinuity at the point where the functions meet, i.e. at x = 3 .


f (x)   = (x 3 -1) = 33 - 1 = 26  
lim x→3+ f (x)   = (x 2 +14) = 32 + 14 = 23  


Since f (x)≠ f (x), f (x) does not exist, and f is not continuous at x = 3.

Problem : Is there a way to define f (c) for the following function so that f (x) is continuous at x = c ?

f (x) = at c = 4    


f (x)   =  
    =  
    = (x + 4)  
    = 8  

Because f (x) = 8 , we should define f (4) = 8 to make this function continuous.

Problem : Is there a way to define f (c) for the following function so that f (x) is continuous at x = c ?

f (x) = at x = 1    


= - ∞ (the numerator approaches 8 and the denominator becomes an extremely small negative number, so the limit goes to - ∞ )

= + ∞ (the numerator approaches 8 and the denominator becomes an extremely small positive number, so the limit goes to + ∞ ) This means that f (x) doesn't exist, so there is no way to define f (1) to make f continuous.

Problem : How can we be sure that the function f (x) = 3x 3 -2x 2 - 31 has a root on the interval [0, 3] ?


We can use the intermediate value theorem. Because f is a polynomial function, it is continuous. f (0) = - 31 and f (3) = 32 . Since 0 lies between f (0) and f (3) , the intermediate value theorem tells us that there is at least one c on [0, 3] for which f (c) = 0 . In other words, f has a root on [0, 3] .

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