Examples

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

Step 1

The Intermediate Value Theorem states that, if $f$ is a real-valued continuous function on the interval $[a, b]$ , and $u$ is a number between $f (a)$ and $f (b)$ , then there is a $c$ contained in the interval $[a, b]$ such that $f (c) = u$ .

$u = f (c) = 0$

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:

$(- \infty, \infty)$

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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