Finite Math Examples

$f (t) = 2 x^{2}$ , $[0, 2]$

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 $f (a) = f (0) = 2 {(0)}^{2}$ .

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Step 3.1

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

$f (0) = 2 \cdot 0$

Step 3.2

Multiply $2$ by $0$ .

$f (0) = 0$

Step 4

Calculate $f (b) = f (2) = 2 {(2)}^{2}$ .

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Step 4.1

Multiply $2$ by ${(2)}^{2}$ by adding the exponents.

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Step 4.1.1

Multiply $2$ by ${(2)}^{2}$ .

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Step 4.1.1.1

Raise $2$ to the power of $1$ .

$f (2) = 2 {(2)}^{2}$

Step 4.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (2) = 2^{1 + 2}$

Step 4.1.2

Add $1$ and $2$ .

$f (2) = 2^{3}$

Step 4.2

Raise $2$ to the power of $3$ .

$f (2) = 8$

Step 5

Since $0$ is on the interval $[0, 8]$ , solve the equation for $x$ at the root by setting $y$ to $0$ in $y = 2 x^{2}$ .

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Step 5.1

Rewrite the equation as $2 x^{2} = 0$ .

$2 x^{2} = 0$

Step 5.2

Divide each term in $2 x^{2} = 0$ by $2$ and simplify.

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Step 5.2.1

Divide each term in $2 x^{2} = 0$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{0}{2}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{0}{2}$

Step 5.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{0}{2}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide $0$ by $2$ .

$x^{2} = 0$

Step 5.3

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

$x = \pm \sqrt{0}$

Step 5.4

Simplify $\pm \sqrt{0}$ .

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Step 5.4.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 5.4.2

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

$x = \pm 0$

Step 5.4.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 6

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[0, 8]$ because $f$ is a continuous function on $[0, 2]$ .

The roots on the interval $[0, 2]$ are located at $x = 0$ .

Step 7