Examples

f (x) = x - 6

$f (x) = x - 6$ ,

(0, 7)

$(0, 7)$

Step 1

The Intermediate Value Theorem states that, if

f

$f$ is a real-valued continuous function on the interval

[a, b]

$[a, b]$ , and

u

$u$ is a number between

f (a)

$f (a)$ and

f (b)

$f (b)$ , then there is a

c

$c$ contained in the interval

[a, b]

$[a, b]$ such that

f (c) = u

$f (c) = u$ .

u = f (c) = 0

$u = f (c) = 0$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

(- \infty, \infty)

$(- \infty, \infty)$

Set-Builder Notation:

{x | x \in ℝ}

${x | x \in ℝ}$

Step 3

Subtract

6

$6$ from

0

$0$ .

f (0) = - 6

$f (0) = - 6$

Step 4

Subtract

6

$6$ from

7

$7$ .

f (7) = 1

$f (7) = 1$

Step 5

Since

0

$0$ is on the interval

[- 6, 1]

$[- 6, 1]$ , solve the equation for

x

$x$ at the root by setting

y

$y$ to

0

$0$ in

y = x - 6

$y = x - 6$ .

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Step 5.1

Rewrite the equation as

x - 6 = 0

$x - 6 = 0$ .

x - 6 = 0

$x - 6 = 0$

Step 5.2

Add

6

$6$ to both sides of the equation.

x = 6

$x = 6$

x = 6

$x = 6$

Step 6

The Intermediate Value Theorem states that there is a root

f (c) = 0

$f (c) = 0$ on the interval

[- 6, 1]

$[- 6, 1]$ because

f

$f$ is a continuous function on

[0, 7]

$[0, 7]$ .

The roots on the interval

[0, 7]

$[0, 7]$ are located at

x = 6

$x = 6$ .

Step 7

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