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].
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