Algebra Examples

$f (x) = x^{3} + 7 x - 2$ , $[0, 10]$

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 $f (a) = f (0) = {(0)}^{3} + 7 (0) - 2$ .

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Step 3.1

Simplify each term.

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Step 3.1.1

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 7 (0) - 2$

Step 3.1.2

Multiply $7$ by $0$ .

$f (0) = 0 + 0 - 2$

Step 3.2

Simplify by adding and subtracting.

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Step 3.2.1

Add $0$ and $0$ .

$f (0) = 0 - 2$

Step 3.2.2

Subtract $2$ from $0$ .

$f (0) = - 2$

Step 4

Calculate $f (b) = f (10) = {(10)}^{3} + 7 (10) - 2$ .

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise $10$ to the power of $3$ .

$f (10) = 1000 + 7 (10) - 2$

Step 4.1.2

Multiply $7$ by $10$ .

$f (10) = 1000 + 70 - 2$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $1000$ and $70$ .

$f (10) = 1070 - 2$

Step 4.2.2

Subtract $2$ from $1070$ .

$f (10) = 1068$

Step 5

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx 0.28249374$

Step 6

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[- 2, 1068]$ because $f$ is a continuous function on $[0, 10]$ .

The roots on the interval $[0, 10]$ are located at $x \approx 0.28249374$ .

Step 7

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