Finite Math Examples

$f (x) = x^{3} + x^{2} - x - 2$ , $[- 2, 1]$

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 $f (a) = f (- 2) = {(- 2)}^{3} + {(- 2)}^{2} - (- 2) - 2$ .

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Step 3.1

Simplify each term.

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Step 3.1.1

Raise $- 2$ to the power of $3$ .

$f (- 2) = - 8 + {(- 2)}^{2} - (- 2) - 2$

Step 3.1.2

Raise $- 2$ to the power of $2$ .

$f (- 2) = - 8 + 4 - (- 2) - 2$

Step 3.1.3

Multiply $- 1$ by $- 2$ .

$f (- 2) = - 8 + 4 + 2 - 2$

Step 3.2

Simplify by adding and subtracting.

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Step 3.2.1

Add $- 8$ and $4$ .

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

Step 3.2.2

Add $- 4$ and $2$ .

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

Step 3.2.3

Subtract $2$ from $- 2$ .

$f (- 2) = - 4$

Step 4

Calculate $f (b) = f (1) = {(1)}^{3} + {(1)}^{2} - (1) - 2$ .

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Step 4.1

Simplify each term.

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Step 4.1.1

One to any power is one.

$f (1) = 1 + {(1)}^{2} - (1) - 2$

Step 4.1.2

One to any power is one.

$f (1) = 1 + 1 - (1) - 2$

Step 4.1.3

Multiply $- 1$ by $1$ .

$f (1) = 1 + 1 - 1 - 2$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $1$ and $1$ .

$f (1) = 2 - 1 - 2$

Step 4.2.2

Subtract $1$ from $2$ .

$f (1) = 1 - 2$

Step 4.2.3

Subtract $2$ from $1$ .

$f (1) = - 1$

Step 5

$0$ is not on the interval $[- 4, - 1]$ .

There is no root on the interval.

Step 6