**Intermediate Value Theorem (IVT):**

Let *f* be continuous on a closed interval [*a*,*b*]. Pick a *y*-value *M* with *f*(*a*)f(*b*) or *f*(*b*)f(*a*). Then there is some *x*-value *c* with *a* < *c* < *b* and *f*(*c*) = *M*. We will use "IVT" interchangeably with Intermediate Value Theorem.

Here's what's going on, in pictures. Start with a continuous function *f* on a closed interval [*a*,*b*]:

Mark on the *y*-axis where *f*(*a*) and *f*(*b*) are:

Pick any value of *M* strictly in between *f*(*a*) and *f*(*b*):

Draw a horizontal dashed line at height *y* = *M*. The IVT guarantees that this dashed line will hit the graph of *f*. In other words, the IVT guarantees the existence of some value *c* strictly in between *a* and *b* where the function value is *M*:

We've said a continuous function is one we can draw without lifting our pen from the paper. The IVT states this more precisely. If a continuous function starts at *f*(*a*) and ends at at *f*(*b*), then as *x* travels from *a* to *b* the function must hit every *y* value in between *f*(*a*) and *f*(*b*):

If a function on [*a,b*] skips a value, that function must be discontinuous:

#### Example 2

Can we use the IVT to conclude that passes through *y* = 1 on ? | |

No. We cannot use the IVT in this case because the function *f* is discontinuous at *x* = 0 and not continuous on | |

#### Example 3

Can we use the IVT to conclude that passes through *y* = 1 on (0,1)? | |

No. This question doesn't even make sense. In order to use the IVT we need to know the function values at the endpoints of the interval, but *f*(0) is undefined.
| |

#### Example 4

Can we use the IVT to conclude that *f*(*x*) =* x*^{2} passes through *y* = 0 on (-1,1)?
| |

No. Since *f*(*a*) = (-1) ^{2} = 1 and *f*(*b*) = (1) ^{2} = 1, 0 is not between *f*(*a*) and *f*(*b*). We cannot use the IVT in this situation. The function *f* does in fact pass through *y* = 0, but we know that by observation, not by using the IVT. Reminder: It doesn't matter whether we have *f*(*a*) < *M* < *f*(*b*) or *f*(*b*) < *M* < *f*(*a*), as long as *M* is between *f*(*a*) and *f*(*b*). | |

#### Exercise 1

- Can we use the IVT to conclude that
*f*(*x*) = *x*^{3} + 2*x* + 1 passes through *y* = 0 on the interval (-2,2)?

Answer

Yes. For starters, *f* is continuous on the closed interval [-2,2]. We have *a* = -2 and *b* = 2, therefore

Since *f*(*a*) = -11 < 0 < 13 = *f*(*b*), the IVT says that there is some *c* with -2 < *c* < 2 and *f*(*c*) = 0.

#### Exercise 2

- Can we use the IVT to conclude that
*f*(*x*) = *e*^{x} passes through y = 0.1 on the interval (0,1)?

Answer

No. *a* = 0 and *b* = 1, therefore *f*(*a*) = 1 and *f*(*b*) = *e*. The value 0.1 is not between 1 and *e*, we can't use the IVT here.

#### Exercise 4

- Can we use the IVT to conclude that
*f*(*x*) = tan(*x*) equals 0 for some *c* in (0,π)?

Answer

No. The function tan(*x*) is not continuous on [0,π] therefore we can't use the IVT.

#### Exercise 5

- Can we use the IVT to conclude that
*f*(*x*) = *x*^{2} passes through 1 on the interval (-1,1)?

Answer

No. The function *f* is indeed continuous on the closed interval [-1,1]. We have *a* = -1 and *b* = 1, therefore
*f*(*a*) = *f*(*b*) = 1. The value 1 is not strictly in between 1 and 1 (if we wrote down 1 < 1 < 1 we would be lying!), we can't use the IVT here.

#### Exercise 6

- Draw a function that is continuous on [0,1] with
*f*(0) = 0, *f*(1) = 1, and *f*(0.5) = 20.

Answer

There are infinitely many right answers to this. Here's one of them:

We could change the above problem to make *f*(0.5) equal anything we want. Therefore

**Be Careful:** We can never use the IVT to conclude that a function *f* *fails* to hit a value *M*.

Another thing to be aware of with the IVT is that it doesn't tell us *where* a function hits a value *M*, or how many times it does so. On its best day, the IVT can only guarantee we hit the value *M* at least once.

#### Exercise 7

- Suppose that
*f* hits every value between *y* = 0 and *y* = 1 on the interval [0,1]. Must *f* be continuous on that interval?

Answer

Answer. No. *f* could look like this, for example:

This function hits every value from 0 to 1 on the interval [0,1], but *f* is discontinuous on that interval.

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