Algebra Examples

$f (x) = x^{3}$ , $[- 4, 4]$

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

Raise $- 4$ to the power of $3$ .

$f (- 4) = - 64$

Step 4

Raise $4$ to the power of $3$ .

$f (4) = 64$

Step 5

Since $0$ is on the interval $[- 64, 64]$ , solve the equation for $x$ at the root by setting $y$ to $0$ in $y = x^{3}$ .

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Step 5.1

Rewrite the equation as $x^{3} = 0$ .

$x^{3} = 0$

Step 5.2

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

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

Step 5.3

Simplify $\sqrt[3]{0}$ .

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Step 5.3.1

Rewrite $0$ as $0^{3}$ .

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

Step 5.3.2

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

$x = 0$

Step 6

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[- 64, 64]$ because $f$ is a continuous function on $[- 4, 4]$ .

The roots on the interval $[- 4, 4]$ are located at $x = 0$ .

Step 7

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