Solved on Feb 11, 2024

Continuous function $f(x)$ has values $\{7, -9, 8, 2, -8\}$ at $x=\{0, 10, 20, 30, 40\}$ . Guarantee roots between $x=0$ and $x=10$ , $x=20$ and $x=30$ , $x=30$ and $x=40$ , but not between $x=10$ and $x=20$ .

STEP 1

Assumptions
1. The function $f(x)$ is continuous everywhere.
2. The table provides specific values of $f(x)$ at $x = 0, 10, 20, 30, 40$ .
3. A root of the function $f(x)$ is a value $x = c$ such that $f(c) = 0$ .
4. The Intermediate Value Theorem states that if $f(x)$ is continuous on the interval $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c$ in $(a, b)$ such that $f(c) = 0$ .

STEP 2

Examine the values of $f(x)$ at $x = 0$ and $x = 10$ .
$f(0) = 7, \quad f(10) = -9$

STEP 3

Since $f(0) = 7$ is positive and $f(10) = -9$ is negative, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 0$ and $x = 10$ .

STEP 4

Examine the values of $f(x)$ at $x = 10$ and $x = 20$ .
$f(10) = -9, \quad f(20) = 8$

STEP 5

Since $f(10) = -9$ is negative and $f(20) = 8$ is positive, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 10$ and $x = 20$ .

STEP 6

Examine the values of $f(x)$ at $x = 20$ and $x = 30$ .
$f(20) = 8, \quad f(30) = 2$

STEP 7

Since both $f(20) = 8$ and $f(30) = 2$ are positive, we cannot guarantee that there is a root between $x = 20$ and $x = 30$ using the Intermediate Value Theorem, as there is no sign change.

STEP 8

Examine the values of $f(x)$ at $x = 30$ and $x = 40$ .
$f(30) = 2, \quad f(40) = -8$

STEP 9

Since $f(30) = 2$ is positive and $f(40) = -8$ is negative, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 30$ and $x = 40$ .

STEP 10

Summarize the guarantees based on the Intermediate Value Theorem:
- Between $x=0$ and $x=10$ , we guarantee there is a root $\checkmark$ - Between $x=10$ and $x=20$ , we guarantee there is a root $\checkmark$ - Between $x=20$ and $x=30$ , we cannot guarantee there is a root since there is no sign change - Between $x=30$ and $x=40$ , we guarantee there is a root $\checkmark$

Was this helpful?

Continuous function $f(x)$ has values $\{7, -9, 8, 2, -8\}$ at $x=\{0, 10, 20, 30, 40\}$ . Guarantee roots between $x=0$ and $x=10$ , $x=20$ and $x=30$ , $x=30$ and $x=40$ , but not between $x=10$ and $x=20$ .

STEP 1

STEP 2

Examine the values of $f(x)$ at $x = 0$ and $x = 10$ .
$f(0) = 7, \quad f(10) = -9$

STEP 3

Since $f(0) = 7$ is positive and $f(10) = -9$ is negative, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 0$ and $x = 10$ .

STEP 4

Examine the values of $f(x)$ at $x = 10$ and $x = 20$ .
$f(10) = -9, \quad f(20) = 8$

STEP 5

Since $f(10) = -9$ is negative and $f(20) = 8$ is positive, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 10$ and $x = 20$ .

STEP 6

Examine the values of $f(x)$ at $x = 20$ and $x = 30$ .
$f(20) = 8, \quad f(30) = 2$

STEP 7

Since both $f(20) = 8$ and $f(30) = 2$ are positive, we cannot guarantee that there is a root between $x = 20$ and $x = 30$ using the Intermediate Value Theorem, as there is no sign change.

STEP 8

Examine the values of $f(x)$ at $x = 30$ and $x = 40$ .
$f(30) = 2, \quad f(40) = -8$

STEP 9

Since $f(30) = 2$ is positive and $f(40) = -8$ is negative, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 30$ and $x = 40$ .

STEP 10

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

STEP 1

STEP 2

Examine the values of f(x)f(x)f(x) at x=0x = 0x=0 and x=10x = 10x=10.f(0)=7,f(10)=−9f(0) = 7, \quad f(10) = -9f(0)=7,f(10)=−9

STEP 3

Since f(0)=7f(0) = 7f(0)=7 is positive and f(10)=−9f(10) = -9f(10)=−9 is negative, and f(x)f(x)f(x) is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between x=0x = 0x=0 and x=10x = 10x=10.

STEP 4

Examine the values of f(x)f(x)f(x) at x=10x = 10x=10 and x=20x = 20x=20.f(10)=−9,f(20)=8f(10) = -9, \quad f(20) = 8f(10)=−9,f(20)=8

STEP 5

Since f(10)=−9f(10) = -9f(10)=−9 is negative and f(20)=8f(20) = 8f(20)=8 is positive, and f(x)f(x)f(x) is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between x=10x = 10x=10 and x=20x = 20x=20.

STEP 6

Examine the values of f(x)f(x)f(x) at x=20x = 20x=20 and x=30x = 30x=30.f(20)=8,f(30)=2f(20) = 8, \quad f(30) = 2f(20)=8,f(30)=2

STEP 7

Since both f(20)=8f(20) = 8f(20)=8 and f(30)=2f(30) = 2f(30)=2 are positive, we cannot guarantee that there is a root between x=20x = 20x=20 and x=30x = 30x=30 using the Intermediate Value Theorem, as there is no sign change.

STEP 8

Examine the values of f(x)f(x)f(x) at x=30x = 30x=30 and x=40x = 40x=40.f(30)=2,f(40)=−8f(30) = 2, \quad f(40) = -8f(30)=2,f(40)=−8

STEP 9

Since f(30)=2f(30) = 2f(30)=2 is positive and f(40)=−8f(40) = -8f(40)=−8 is negative, and f(x)f(x)f(x) is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between x=30x = 30x=30 and x=40x = 40x=40.

STEP 10

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Examine the values of $f(x)$ at $x = 0$ and $x = 10$ .
$f(0) = 7, \quad f(10) = -9$

Since $f(0) = 7$ is positive and $f(10) = -9$ is negative, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 0$ and $x = 10$ .

Examine the values of $f(x)$ at $x = 10$ and $x = 20$ .
$f(10) = -9, \quad f(20) = 8$

Since $f(10) = -9$ is negative and $f(20) = 8$ is positive, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 10$ and $x = 20$ .

Examine the values of $f(x)$ at $x = 20$ and $x = 30$ .
$f(20) = 8, \quad f(30) = 2$

Since both $f(20) = 8$ and $f(30) = 2$ are positive, we cannot guarantee that there is a root between $x = 20$ and $x = 30$ using the Intermediate Value Theorem, as there is no sign change.

Examine the values of $f(x)$ at $x = 30$ and $x = 40$ .
$f(30) = 2, \quad f(40) = -8$

Since $f(30) = 2$ is positive and $f(40) = -8$ is negative, and $f(x)$ is continuous everywhere, by the Intermediate Value Theorem, there must be at least one root between $x = 30$ and $x = 40$ .