Finite Math Examples

$f (x) = 4 x - \frac{1}{2}$ , $(1, 3)$

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 $f (a) = f (1) = 4 (1) - \frac{1}{2}$ .

Tap for more steps...

Step 3.1

Multiply $4$ by $1$ .

$f (1) = 4 - \frac{1}{2}$

Step 3.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (1) = 4 \cdot \frac{2}{2} - \frac{1}{2}$

Step 3.3

Combine $4$ and $\frac{2}{2}$ .

$f (1) = \frac{4 \cdot 2}{2} - \frac{1}{2}$

Step 3.4

Combine the numerators over the common denominator.

$f (1) = \frac{4 \cdot 2 - 1}{2}$

Step 3.5

Simplify the numerator.

Tap for more steps...

Step 3.5.1

Multiply $4$ by $2$ .

$f (1) = \frac{8 - 1}{2}$

Step 3.5.2

Subtract $1$ from $8$ .

$f (1) = \frac{7}{2}$

Step 4

Calculate $f (b) = f (3) = 4 (3) - \frac{1}{2}$ .

Tap for more steps...

Step 4.1

Multiply $4$ by $3$ .

$f (3) = 12 - \frac{1}{2}$

Step 4.2

To write $12$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (3) = 12 \cdot \frac{2}{2} - \frac{1}{2}$

Step 4.3

Combine $12$ and $\frac{2}{2}$ .

$f (3) = \frac{12 \cdot 2}{2} - \frac{1}{2}$

Step 4.4

Combine the numerators over the common denominator.

$f (3) = \frac{12 \cdot 2 - 1}{2}$

Step 4.5

Simplify the numerator.

Tap for more steps...

Step 4.5.1

Multiply $12$ by $2$ .

$f (3) = \frac{24 - 1}{2}$

Step 4.5.2

Subtract $1$ from $24$ .

$f (3) = \frac{23}{2}$

Step 5

$0$ is not on the interval $[\frac{7}{2}, \frac{23}{2}]$ .

There is no root on the interval.

Step 6