Finite Math Examples

f (x) = x^{3}

$f (x) = x^{3}$ ,

[- 4, 4]

$[- 4, 4]$

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

Raise

- 4

$- 4$ to the power of

3

$3$ .

f (- 4) = - 64

$f (- 4) = - 64$

Step 4

Raise

4

$4$ to the power of

3

$3$ .

f (4) = 64

$f (4) = 64$

Step 5

Since

0

$0$ is on the interval

[- 64, 64]

$[- 64, 64]$ , solve the equation for

x

$x$ at the root by setting

y

$y$ to

0

$0$ in

y = x^{3}

$y = x^{3}$ .

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Step 5.1

Rewrite the equation as

x^{3} = 0

$x^{3} = 0$ .

x^{3} = 0

$x^{3} = 0$

Step 5.2

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

x = \sqrt[3]{0}

$x = \sqrt[3]{0}$

Step 5.3

Simplify

\sqrt[3]{0}

$\sqrt[3]{0}$ .

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Step 5.3.1

Rewrite

0

$0$ as

0^{3}

$0^{3}$ .

x = \sqrt[3]{0^{3}}

$x = \sqrt[3]{0^{3}}$

Step 5.3.2

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

x = 0

$x = 0$

x = 0

$x = 0$

x = 0

$x = 0$

Step 6

The Intermediate Value Theorem states that there is a root

f (c) = 0

$f (c) = 0$ on the interval

[- 64, 64]

$[- 64, 64]$ because

f

$f$ is a continuous function on

[- 4, 4]

$[- 4, 4]$ .

The roots on the interval

[- 4, 4]

$[- 4, 4]$ are located at

x = 0

$x = 0$ .

Step 7

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