Finite Math Examples

$y = 64 - x^{2}$ , $[- 8, 8]$

Step 1

Reorder $64$ and $- x^{2}$ .

$y = - x^{2} + 64$

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 $f (a) = f (- 8) = - {(- 8)}^{2} + 64$ .

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise $- 8$ to the power of $2$ .

$f (- 8) = - 1 \cdot 64 + 64$

Step 4.1.2

Multiply $- 1$ by $64$ .

$f (- 8) = - 64 + 64$

Step 4.2

Add $- 64$ and $64$ .

$f (- 8) = 0$

Step 5

Calculate $f (b) = f (8) = - {(8)}^{2} + 64$ .

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Step 5.1

Simplify each term.

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Step 5.1.1

Raise $8$ to the power of $2$ .

$f (8) = - 1 \cdot 64 + 64$

Step 5.1.2

Multiply $- 1$ by $64$ .

$f (8) = - 64 + 64$

Step 5.2

Add $- 64$ and $64$ .

$f (8) = 0$

Step 6

Since $0$ is on the interval $[0, 0]$ , solve the equation for $x$ at the root by setting $y$ to $0$ in $y = - x^{2} + 64$ .

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Step 6.1

Rewrite the equation as $- x^{2} + 64 = 0$ .

$- x^{2} + 64 = 0$

Step 6.2

Subtract $64$ from both sides of the equation.

$- x^{2} = - 64$

Step 6.3

Divide each term in $- x^{2} = - 64$ by $- 1$ and simplify.

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Step 6.3.1

Divide each term in $- x^{2} = - 64$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 64}{- 1}$

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.

$\frac{x^{2}}{1} = \frac{- 64}{- 1}$

Step 6.3.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 64}{- 1}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Divide $- 64$ by $- 1$ .

$x^{2} = 64$

Step 6.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{64}$

Step 6.5

Simplify $\pm \sqrt{64}$ .

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Step 6.5.1

Rewrite $64$ as $8^{2}$ .

$x = \pm \sqrt{8^{2}}$

Step 6.5.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 8$

Step 6.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 8$

Step 6.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 8$

Step 6.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 8, - 8$

Step 7

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[0, 0]$ because $f$ is a continuous function on $[- 8, 8]$ .

The roots on the interval $[- 8, 8]$ are located at $x = 8, x = - 8$ .

Step 8