Finite Math Examples

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

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

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Step 3.1

Remove parentheses.

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

Step 3.2

Raise $- 1$ to the power of $2$ .

$f (- 1) = 1 - 1$

Step 3.3

Subtract $1$ from $1$ .

$f (- 1) = 0$

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

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

$f (2) = 4 + 2$

Step 4.3

Add $4$ and $2$ .

$f (2) = 6$

Step 5

Since $0$ is on the interval $[0, 6]$ , solve the equation for $x$ at the root by setting $y$ to $0$ in $y = x^{2} + x$ .

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Step 5.1

Rewrite the equation as $x^{2} + x = 0$ .

$x^{2} + x = 0$

Step 5.2

Factor $x$ out of $x^{2} + x$ .

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Step 5.2.1

Factor $x$ out of $x^{2}$ .

$x \cdot x + x = 0$

Step 5.2.2

Raise $x$ to the power of $1$ .

$x \cdot x + x = 0$

Step 5.2.3

Factor $x$ out of $x^{1}$ .

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

Step 5.2.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$x (x + 1) = 0$

Step 5.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 1 = 0$

Step 5.4

Set $x$ equal to $0$ .

$x = 0$

Step 5.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 5.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 5.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.6

The final solution is all the values that make $x (x + 1) = 0$ true.

$x = 0, - 1$

Step 6

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[0, 6]$ because $f$ is a continuous function on $[- 1, 2]$ .

The roots on the interval $[- 1, 2]$ are located at $x = 0, x = - 1$ .

Step 7