Finite Math Examples

$f (n) = 2 {(\frac{1}{3})}^{n - 1} + 4$ , $h (t) = \frac{t + 4}{t - 2}$ , $((h \circ f)) (0)$

Step 1

Set up the composite result function.

$h (f (n))$

Step 2

Evaluate $h (2 {(\frac{1}{3})}^{n - 1} + 4)$ by substituting in the value of $f$ into $h$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{(2 {(\frac{1}{3})}^{n - 1} + 4) + 4}{(2 {(\frac{1}{3})}^{n - 1} + 4) - 2}$

Step 3

Cancel the common factor of $(2 {(\frac{1}{3})}^{n - 1} + 4) + 4$ and $(2 {(\frac{1}{3})}^{n - 1} + 4) - 2$ .

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Step 3.1

Factor $2$ out of $2 {(\frac{1}{3})}^{n - 1}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1}) + 4 + 4}{(2 {(\frac{1}{3})}^{n - 1} + 4) - 2}$

Step 3.2

Factor $2$ out of $4$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1}) + 2 (2) + 4}{(2 {(\frac{1}{3})}^{n - 1} + 4) - 2}$

Step 3.3

Factor $2$ out of $2 ({(\frac{1}{3})}^{n - 1}) + 2 (2)$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2) + 4}{(2 {(\frac{1}{3})}^{n - 1} + 4) - 2}$

Step 3.4

Factor $2$ out of $4$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2) + 2 (2)}{(2 {(\frac{1}{3})}^{n - 1} + 4) - 2}$

Step 3.5

Factor $2$ out of $2 ({(\frac{1}{3})}^{n - 1} + 2) + 2 (2)$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{(2 {(\frac{1}{3})}^{n - 1} + 4) - 2}$

Step 3.6

Cancel the common factors.

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Step 3.6.1

Factor $2$ out of $2 {(\frac{1}{3})}^{n - 1}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{2 ({(\frac{1}{3})}^{n - 1}) + 4 - 2}$

Step 3.6.2

Factor $2$ out of $4$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{2 ({(\frac{1}{3})}^{n - 1}) + 2 (2) - 2}$

Step 3.6.3

Factor $2$ out of $2 ({(\frac{1}{3})}^{n - 1}) + 2 (2)$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{2 ({(\frac{1}{3})}^{n - 1} + 2) - 2}$

Step 3.6.4

Factor $2$ out of $- 2$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{2 ({(\frac{1}{3})}^{n - 1} + 2) + 2 (- 1)}$

Step 3.6.5

Factor $2$ out of $2 ({(\frac{1}{3})}^{n - 1} + 2) + 2 (- 1)$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{2 ({(\frac{1}{3})}^{n - 1} + 2 - 1)}$

Step 3.6.6

Cancel the common factor.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{2 ({(\frac{1}{3})}^{n - 1} + 2 + 2)}{2 ({(\frac{1}{3})}^{n - 1} + 2 - 1)}$

Step 3.6.7

Rewrite the expression.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{{(\frac{1}{3})}^{n - 1} + 2 + 2}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 4

Simplify the numerator.

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Step 4.1

Apply the product rule to $\frac{1}{3}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1^{n - 1}}{3^{n - 1}} + 2 + 2}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 4.2

One to any power is one.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1}{3^{n - 1}} + 2 + 2}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 4.3

Add $2$ and $2$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1}{3^{n - 1}} + 4}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 4.4

To write $4$ as a fraction with a common denominator, multiply by $\frac{3^{n - 1}}{3^{n - 1}}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1}{3^{n - 1}} + 4 \cdot \frac{3^{n - 1}}{3^{n - 1}}}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 4.5

Combine $4$ and $\frac{3^{n - 1}}{3^{n - 1}}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1}{3^{n - 1}} + \frac{4 \cdot 3^{n - 1}}{3^{n - 1}}}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 4.6

Combine the numerators over the common denominator.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}}}{{(\frac{1}{3})}^{n - 1} + 2 - 1}$

Step 5

Simplify the denominator.

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Step 5.1

Apply the product rule to $\frac{1}{3}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}}}{\frac{1^{n - 1}}{3^{n - 1}} + 2 - 1}$

Step 5.2

One to any power is one.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}}}{\frac{1}{3^{n - 1}} + 2 - 1}$

Step 5.3

Subtract $1$ from $2$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}}}{\frac{1}{3^{n - 1}} + 1}$

Step 5.4

Write $1$ as a fraction with a common denominator.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}}}{\frac{1}{3^{n - 1}} + \frac{3^{n - 1}}{3^{n - 1}}}$

Step 5.5

Combine the numerators over the common denominator.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{\frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}}}{\frac{1 + 3^{n - 1}}{3^{n - 1}}}$

Step 6

Multiply the numerator by the reciprocal of the denominator.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}} \cdot \frac{3^{n - 1}}{1 + 3^{n - 1}}$

Step 7

Cancel the common factor of $3^{n - 1}$ .

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Step 7.1

Cancel the common factor.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{1 + 4 \cdot 3^{n - 1}}{3^{n - 1}} \cdot \frac{3^{n - 1}}{1 + 3^{n - 1}}$

Step 7.2

Rewrite the expression.

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = (1 + 4 \cdot 3^{n - 1}) (\frac{1}{1 + 3^{n - 1}})$

Step 8

Multiply $1 + 4 \cdot 3^{n - 1}$ by $\frac{1}{1 + 3^{n - 1}}$ .

$h (2 {(\frac{1}{3})}^{n - 1} + 4) = \frac{1 + 4 \cdot 3^{n - 1}}{1 + 3^{n - 1}}$

Step 9

Evaluate the result function by replacing $n$ with $0$ in the function.

$\frac{1 + 4 \cdot 3^{(0) - 1}}{1 + 3^{(0) - 1}}$

Step 10

Simplify the evaluated function.

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Step 10.1

Simplify the numerator.

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Step 10.1.1

Subtract $1$ from $0$ .

$\frac{1 + 4 \cdot 3^{- 1}}{1 + 3^{0 - 1}}$

Step 10.1.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1 + 4 \cdot \frac{1}{3}}{1 + 3^{0 - 1}}$

Step 10.1.3

Combine $4$ and $\frac{1}{3}$ .

$\frac{1 + \frac{4}{3}}{1 + 3^{0 - 1}}$

Step 10.1.4

Write $1$ as a fraction with a common denominator.

$\frac{\frac{3}{3} + \frac{4}{3}}{1 + 3^{0 - 1}}$

Step 10.1.5

Combine the numerators over the common denominator.

$\frac{\frac{3 + 4}{3}}{1 + 3^{0 - 1}}$

Step 10.1.6

Add $3$ and $4$ .

$\frac{\frac{7}{3}}{1 + 3^{0 - 1}}$

Step 10.2

Simplify the denominator.

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Step 10.2.1

Subtract $1$ from $0$ .

$\frac{\frac{7}{3}}{1 + 3^{- 1}}$

Step 10.2.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{\frac{7}{3}}{1 + \frac{1}{3}}$

Step 10.2.3

Write $1$ as a fraction with a common denominator.

$\frac{\frac{7}{3}}{\frac{3}{3} + \frac{1}{3}}$

Step 10.2.4

Combine the numerators over the common denominator.

$\frac{\frac{7}{3}}{\frac{3 + 1}{3}}$

Step 10.2.5

Add $3$ and $1$ .

$\frac{\frac{7}{3}}{\frac{4}{3}}$

Step 10.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{7}{3} \cdot \frac{3}{4}$

Step 10.4

Cancel the common factor of $3$ .

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Step 10.4.1

Cancel the common factor.

$\frac{7}{3} \cdot \frac{3}{4}$

Step 10.4.2

Rewrite the expression.

$7 (\frac{1}{4})$

Step 10.5

Combine $7$ and $\frac{1}{4}$ .

$\frac{7}{4}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\frac{7}{4}$

Decimal Form:

$1.75$

Mixed Number Form:

$1 \frac{3}{4}$