Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| -2 | 2 |
| 1 | -1 |
| 3 | -1 |
| 4 | 5 |
Statement: "f has exactly one x-intercept".
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
- On the domain -2 \leq x \leq 1, f\left(x\right) takes every value between 2 and -1. Therefore, f\left(x\right) = 0 for some x on the domain -2 \leq x \leq 1.
- On the domain 3 \leq x \leq 4 , f\left(x\right) takes every value between -1 and 5. Therefore, f\left(x\right) = 0 for some x on the domain 3 \leq x \leq 4.
There are therefore at least two x-intercepts.
The statement "f has exactly one x-intercept" is false since we know it has at least two x-intercepts.
Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| 2 | -2 |
| 3 | -2 |
| 4 | 1 |
| 5 | -4 |
| 6 | -3 |
| 7 | 5 |
Statement: "f has at least two x-intercepts on the domain 3 \leq x \leq 5 "
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
- On the domain 3 \leq x \leq 4, f\left(x\right) takes every value between -2 and 1. Therefore, f\left(x\right) = 0 for some x on the domain 3 \lt x \lt 4.
- On the domain 4 \leq x \leq 5, f\left(x\right) takes every value between 1 and -4. Therefore, f\left(x\right) = 0 for some x on the domain 4 \lt x \lt 5.
There are therefore at least two x-intercepts on the domain 3 \leq x \leq 5.
The statement "f has at least two x-intercepts on the domain 3 \leq x \leq 5 " is true.
Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| -1 | 1 |
| 0 | 1 |
| 1 | -1 |
| 4 | 7 |
| 6 | 7 |
| 7 | -2 |
Statement: "f has exactly two x-intercept".
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
- On the domain 0 \leq x \leq 1, f\left(x\right) takes every value between 1 and -1. Therefore, f\left(x\right) = 0 for some x on the domain 0 \lt x \lt 1.
- On the domain 1 \leq x \leq 4, f\left(x\right) takes every value between -1 and 7. Therefore, f\left(x\right) = 0 for some x on the domain 1 \lt x \lt 4
- On the domain 6 \leq x \leq 7, f\left(x\right) takes every value between 7 and -2. Therefore, f\left(x\right) = 0 for some x on the domain 6 \lt x \lt 7
There are therefore at least three x-intercepts.
The statement "f has exactly two x-intercepts" is false since we know it has at least three x-intercepts.
Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| -3 | 9 |
| 0 | 7 |
| 2 | 9 |
| 5 | -4 |
| 6 | -3 |
| 8 | -4 |
Statement: "f has at least one x-intercept."
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
On the domain 2 \leq x \leq 5, f\left(x\right) takes every value between 9 and -4. Therefore, f\left(x\right) = 0 for some x on the domain 2 \lt x \lt 5.
There is therefore at least one x-intercept.
The statement "f has exactly one x-intercept" is true.
Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| -1 | 1 |
| 1 | -1 |
| 3 | 3 |
| 5 | 4 |
| 8 | 3 |
| 13 | 2 |
| 14 | -2 |
Statement: "f has no more than two x-intercepts."
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
- On the domain -1 \leq x \leq 1, f\left(x\right) takes every value between 1 and -1. Therefore, f\left(x\right) = 0 for some x on the domain -1 \lt x \lt 1
- On the domain 1 \leq x \leq 3, f\left(x\right) takes every value between -1 and 3. Therefore, f\left(x\right) = 0 for some x on the domain 1 \lt x \lt 3
- On the domain 13 \leq x \leq 14, f\left(x\right) takes every value between 2 and -2. Therefore, f\left(x\right) = 0 for some x on the domain 13 \lt x \lt 14
There are therefore at least three x-intercepts.
The statement "f has no more than two x-intercepts" is false since we know it has at least three x-intercepts.
Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| -3 | -4 |
| 3 | -2 |
| 4 | 2 |
| 5 | 8 |
| 8 | -2 |
Statement: "f has no x-intercepts on the domain -3 \leq x \leq 4. "
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
On the domain -3 \leq x \leq 4, f\left(x\right) takes every value between -4 and 2. Therefore, f\left(x\right) = 0 for some x on the domain -3 \lt x \lt 4.
There is therefore at least one x-intercept.
The statement "f has no x-intercepts on the domain - 3 \leq x \leq 4 " is false since we know it has at least one x-intercept on that domain.
Here are some values of a function f, that is known to be continuous.
| x | f(x) |
|---|---|
| -10 | 4 |
| -4 | -2 |
| -3 | -2 |
| -1 | -5 |
| 4 | 3 |
| 5 | 3 |
| 7 | 6 |
Statement: "f has less than two x-intercepts."
The intermediate value theorem states that if a continuous function f, with an interval \left[a, b\right] as its domain, takes values f\left(a\right) and f\left(b\right) at each end of the interval, then it also takes any value between f\left(a\right) and f\left(b\right) at some point within the interval.
Applying the intermediate value theorem to the data here leads to the following conclusions:
- On the domain -10 \leq x \leq -4, f\left(x\right) takes every value between 4 and -2. Therefore, f\left(x\right) = 0 for some x on the domain -10 \lt x \lt -4
- On the domain -1 \leq x \leq 4, f\left(x\right) takes every value between -5 and 3. Therefore, f\left(x\right) = 0 for some x on the domain -1 \lt x \lt 4
There are therefore at least two x-intercepts.
The statement "f has less than two x-intercepts" is false since we know it has at least two x-intercepts.