Examples

f (x) = x^{3} + 7 x - 2

$f (x) = x^{3} + 7 x - 2$ ,

[0, 10]

$[0, 10]$

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:

${x | x \in ℝ}$

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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