Finite Math Examples

f (x) = x^{3} + x^{2} - x - 2

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

[- 2, 1]

$[- 2, 1]$

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

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

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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