Examples

f (x) = - x^{2} + x

$f (x) = - x^{2} + x$ ,

[- 2, 2]

$[- 2, 2]$

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:

${x | x \in ℝ}$

Step 3

Calculate

$f (a) = f (- 2) = - {(- 2)}^{2} - 2$ .

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Step 3.1

Remove parentheses.

$f (- 2) = - {(- 2)}^{2} - 2$

Step 3.2

Simplify each term.

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Step 3.2.1

Raise

$- 2$ to the power of

$2$ .

$f (- 2) = - 1 \cdot 4 - 2$

Step 3.2.2

Multiply

$- 1$ by

$4$ .

$f (- 2) = - 4 - 2$

Step 3.3

Subtract

$2$ from

$- 4$ .

$f (- 2) = - 6$

Step 4

Calculate

$f (b) = f (2) = - {(2)}^{2} + 2$ .

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Step 4.1

Remove parentheses.

$f (2) = - {(2)}^{2} + 2$

Step 4.2

Simplify each term.

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Step 4.2.1

Raise

$2$ to the power of

$2$ .

$f (2) = - 1 \cdot 4 + 2$

Step 4.2.2

Multiply

$- 1$ by

$4$ .

$f (2) = - 4 + 2$

Step 4.3

Add

$- 4$ and

$2$ .

$f (2) = - 2$

Step 5

$0$ is not on the interval

$[- 6, - 2]$ .

There is no root on the interval.

Step 6

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