**Problem : **
Is *f* (*x*) = *x*^{3} + 2*x* + 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.

**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

f (x) | = x + 3 = 2 + 3 = 5 | ||

f (x) | = (x^{4} -11) = 2^{4} - 11 = 5 |

Thus,

**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

f (x) | = (x^{3} -1) = 3^{3} - 1 = 26 | ||

lim_{x→3+}f (x) | = (x^{2} +14) = 3^{2} + 14 = 23 |

Since

**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

**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

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

We can use the intermediate value theorem. Because

Take a Study Break!