Finite Math Examples

$(- 5, 5)$ , $x = 4$

Step 1

Subtract $x$ from both sides of the equation.

$0 = 4 - x$

Step 2

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 3

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 4

Calculate $f (a) = f (- 5) = 4 - (- 5)$ .

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Step 4.1

Multiply $- 1$ by $- 5$ .

$f (- 5) = 4 + 5$

Step 4.2

Add $4$ and $5$ .

$f (- 5) = 9$

Step 5

Calculate $f (b) = f (5) = 4 - (5)$ .

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Step 5.1

Multiply $- 1$ by $5$ .

$f (5) = 4 - 5$

Step 5.2

Subtract $5$ from $4$ .

$f (5) = - 1$

Step 6

Since $0$ is on the interval $[- 1, 9]$ , solve the equation for $x$ at the root.

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Step 6.1

Rewrite the equation as $4 - x = 0$ .

$4 - x = 0$

Step 6.2

Subtract $4$ from both sides of the equation.

$- x = - 4$

Step 6.3

Divide each term in $- x = - 4$ by $- 1$ and simplify.

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Step 6.3.1

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Step 6.3.2.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Divide $- 4$ by $- 1$ .

$x = 4$

Step 7

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[- 1, 9]$ because $f$ is a continuous function on $[- 5, 5]$ .

The roots on the interval $[- 5, 5]$ are located at $x = 4$ .

Step 8