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Problems for "Continuity"

Problems for "Continuity"

Problems for "Continuity"

Problems for "Continuity"

Problem : Is f (x) = x3 + 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) = (x4 -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) = (x3 -1) = 33 - 1 = 26  
limx→3+f (x) = (x2 +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) = 3x3 -2x2 - 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].