Finite Math Examples

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

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

[- 1, 2]

$[- 1, 2]$

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}

${x | x \in ℝ}$

Step 3

Calculate

f (a) = f (- 1) = {(- 1)}^{2} - 1

$f (a) = f (- 1) = {(- 1)}^{2} - 1$ .

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Step 3.1

Remove parentheses.

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

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

Step 3.2

Raise

- 1

$- 1$ to the power of

2

$2$ .

f (- 1) = 1 - 1

$f (- 1) = 1 - 1$

Step 3.3

Subtract

1

$1$ from

1

$1$ .

f (- 1) = 0

$f (- 1) = 0$

f (- 1) = 0

$f (- 1) = 0$

Step 4

Calculate

f (b) = f (2) = {(2)}^{2} + 2

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

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Step 4.1

Remove parentheses.

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

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

Step 4.2

Raise

2

$2$ to the power of

2

$2$ .

f (2) = 4 + 2

$f (2) = 4 + 2$

Step 4.3

Add

4

$4$ and

2

$2$ .

f (2) = 6

$f (2) = 6$

f (2) = 6

$f (2) = 6$

Step 5

Since

0

$0$ is on the interval

[0, 6]

$[0, 6]$ , solve the equation for

x

$x$ at the root by setting

y

$y$ to

0

$0$ in

y = x^{2} + x

$y = x^{2} + x$ .

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Step 5.1

Rewrite the equation as

x^{2} + x = 0

$x^{2} + x = 0$ .

x^{2} + x = 0

$x^{2} + x = 0$

Step 5.2

Factor

x

$x$ out of

x^{2} + x

$x^{2} + x$ .

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Step 5.2.1

Factor

x

$x$ out of

x^{2}

$x^{2}$ .

x \cdot x + x = 0

$x \cdot x + x = 0$

Step 5.2.2

Raise

x

$x$ to the power of

1

$1$ .

x \cdot x + x = 0

$x \cdot x + x = 0$

Step 5.2.3

Factor

x

$x$ out of

x^{1}

$x^{1}$ .

x \cdot x + x \cdot 1 = 0

$x \cdot x + x \cdot 1 = 0$

Step 5.2.4

Factor

x

$x$ out of

x \cdot x + x \cdot 1

$x \cdot x + x \cdot 1$ .

x (x + 1) = 0

$x (x + 1) = 0$

x (x + 1) = 0

$x (x + 1) = 0$

Step 5.3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x = 0

$x = 0$

x + 1 = 0

$x + 1 = 0$

Step 5.4

Set

x

$x$ equal to

0

$0$ .

x = 0

$x = 0$

Step 5.5

Set

x + 1

$x + 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.5.1

Set

x + 1

$x + 1$ equal to

0

$0$ .

x + 1 = 0

$x + 1 = 0$

Step 5.5.2

Subtract

1

$1$ from both sides of the equation.

x = - 1

$x = - 1$

x = - 1

$x = - 1$

Step 5.6

The final solution is all the values that make

x (x + 1) = 0

$x (x + 1) = 0$ true.

x = 0, - 1

$x = 0, - 1$

x = 0, - 1

$x = 0, - 1$

Step 6

The Intermediate Value Theorem states that there is a root

f (c) = 0

$f (c) = 0$ on the interval

[0, 6]

$[0, 6]$ because

f

$f$ is a continuous function on

[- 1, 2]

$[- 1, 2]$ .

The roots on the interval

[- 1, 2]

$[- 1, 2]$ are located at

x = 0, x = - 1

$x = 0, x = - 1$ .

Step 7