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

Solution for Problem 1 >>

Yes. All polynomial functions are continuous. To check,

f (x) = 13 and f (2) = 13

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

Problem : Is the following function continuous?

f (x) =

Solution for Problem 3 >>

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.

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Problem : Is the following function continuous?

f (x) =

Solution for Problem 4 >>

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.

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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

Solution for Problem 5 >>

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

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

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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

Solution for Problem 6 >>

= - ∞ (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.

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