Precalculus Examples

$A = 1100 {(1 + i)}^{3}$

Step 1

Write $A = 1100 {(1 + i)}^{3}$ as a function.

$f (x) = 1100 {(1 + i)}^{3}$

Step 2

Consider the difference quotient formula.

$\frac{f (x + h) - f (x)}{h}$

Step 3

Find the components of the definition.

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Step 3.1

Evaluate the function at $x = x + h$ .

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Step 3.1.1

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = 1100 {(1 + i)}^{3}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Use the Binomial Theorem.

$f (x + h) = 1100 (1^{3} + 3 \cdot (1^{2} i) + 3 \cdot (1 i^{2}) + i^{3})$

Step 3.1.2.2

Simplify terms.

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Step 3.1.2.2.1

Simplify each term.

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Step 3.1.2.2.1.1

One to any power is one.

$f (x + h) = 1100 (1 + 3 \cdot (1^{2} i) + 3 \cdot (1 i^{2}) + i^{3})$

Step 3.1.2.2.1.2

One to any power is one.

$f (x + h) = 1100 (1 + 3 \cdot (1 i) + 3 \cdot (1 i^{2}) + i^{3})$

Step 3.1.2.2.1.3

Multiply $3$ by $1$ .

$f (x + h) = 1100 (1 + 3 i + 3 \cdot (1 i^{2}) + i^{3})$

Step 3.1.2.2.1.4

Multiply $3$ by $1$ .

$f (x + h) = 1100 (1 + 3 i + 3 i^{2} + i^{3})$

Step 3.1.2.2.1.5

Rewrite $i^{2}$ as $- 1$ .

$f (x + h) = 1100 (1 + 3 i + 3 \cdot - 1 + i^{3})$

Step 3.1.2.2.1.6

Multiply $3$ by $- 1$ .

$f (x + h) = 1100 (1 + 3 i - 3 + i^{3})$

Step 3.1.2.2.1.7

Factor out $i^{2}$ .

$f (x + h) = 1100 (1 + 3 i - 3 + i^{2} \cdot i)$

Step 3.1.2.2.1.8

Rewrite $i^{2}$ as $- 1$ .

$f (x + h) = 1100 (1 + 3 i - 3 - 1 \cdot i)$

Step 3.1.2.2.1.9

Rewrite $- 1 i$ as $- i$ .

$f (x + h) = 1100 (1 + 3 i - 3 - i)$

Step 3.1.2.2.2

Simplify terms.

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Step 3.1.2.2.2.1

Subtract $3$ from $1$ .

$f (x + h) = 1100 (- 2 + 3 i - i)$

Step 3.1.2.2.2.2

Subtract $i$ from $3 i$ .

$f (x + h) = 1100 (- 2 + 2 i)$

Step 3.1.2.2.2.3

Apply the distributive property.

$f (x + h) = 1100 \cdot - 2 + 1100 (2 i)$

Step 3.1.2.2.2.4

Multiply.

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Step 3.1.2.2.2.4.1

Multiply $1100$ by $- 2$ .

$f (x + h) = - 2200 + 1100 (2 i)$

Step 3.1.2.2.2.4.2

Multiply $2$ by $1100$ .

$f (x + h) = - 2200 + 2200 i$

Step 3.1.2.3

The final answer is $- 2200 + 2200 i$ .

$- 2200 + 2200 i$

Step 3.2

Find the components of the definition.

$f (x + h) = - 2200 + 2200 i$

$f (x) = - 2200 + 2200 i$

$f (x + h) = - 2200 + 2200 i$

$f (x) = - 2200 + 2200 i$

Step 4

Plug in the components.

$\frac{f (x + h) - f (x)}{h} = \frac{- 2200 + 2200 i - (- 2200 + 2200 i)}{h}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Apply the distributive property.

$\frac{- 2200 + 2200 i - - 2200 - (2200 i)}{h}$

Step 5.1.2

Multiply $- 1$ by $- 2200$ .

$\frac{- 2200 + 2200 i + 2200 - (2200 i)}{h}$

Step 5.1.3

Multiply $2200$ by $- 1$ .

$\frac{- 2200 + 2200 i + 2200 - 2200 i}{h}$

Step 5.1.4

Add $- 2200$ and $2200$ .

$\frac{0 + 2200 i - 2200 i}{h}$

Step 5.1.5

Add $0$ and $2200 i$ .

$\frac{2200 i - 2200 i}{h}$

Step 5.1.6

Subtract $2200 i$ from $2200 i$ .

$\frac{0}{h}$

Step 5.2

Divide $0$ by $h$ .

$0$

Step 6