Precalculus Examples

$y = {(1.06)}^{8 t}$

Step 1

Write $y = {(1.06)}^{8 t}$ as a function.

$f (t) = {(1.06)}^{8 t}$

Step 2

Consider the difference quotient formula.

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

Step 3

Find the components of the definition.

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Step 3.1

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

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Step 3.1.1

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

$f (t + h) = {(1.06)}^{8 (t + h)}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Apply the distributive property.

$f (t + h) = {1.06}^{8 t + 8 h}$

Step 3.1.2.2

The final answer is ${1.06}^{8 t + 8 h}$ .

${1.06}^{8 t + 8 h}$

Step 3.2

Find the components of the definition.

$f (t + h) = {1.06}^{8 t + 8 h}$

$f (t) = {1.06}^{8 t}$

$f (t + h) = {1.06}^{8 t + 8 h}$

$f (t) = {1.06}^{8 t}$

Step 4

Plug in the components.

$\frac{f (t + h) - f t}{h} = \frac{{1.06}^{8 t + 8 h} - ({1.06}^{8 t})}{h}$

Step 5

Simplify the numerator.

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Step 5.1

Rewrite ${1.06}^{8 t + 8 h}$ as ${({1.06}^{4 t + 4 h})}^{2}$ .

$\frac{{({1.06}^{4 t + 4 h})}^{2} - {1.06}^{8 t}}{h}$

Step 5.2

Rewrite ${1.06}^{8 t}$ as ${({1.06}^{4 t})}^{2}$ .

$\frac{{({1.06}^{4 t + 4 h})}^{2} - {({1.06}^{4 t})}^{2}}{h}$

Step 5.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = {1.06}^{4 t + 4 h}$ and $b = {1.06}^{4 t}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) ({1.06}^{4 t + 4 h} - {1.06}^{4 t})}{h}$

Step 5.4

Simplify.

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Step 5.4.1

Rewrite ${1.06}^{4 t + 4 h}$ as ${({1.06}^{2 t + 2 h})}^{2}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) ({({1.06}^{2 t + 2 h})}^{2} - {1.06}^{4 t})}{h}$

Step 5.4.2

Rewrite ${1.06}^{4 t}$ as ${({1.06}^{2 t})}^{2}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) ({({1.06}^{2 t + 2 h})}^{2} - {({1.06}^{2 t})}^{2})}{h}$

Step 5.4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = {1.06}^{2 t + 2 h}$ and $b = {1.06}^{2 t}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) (({1.06}^{2 t + 2 h} + {1.06}^{2 t}) ({1.06}^{2 t + 2 h} - {1.06}^{2 t}))}{h}$

Step 5.4.4

Simplify.

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Step 5.4.4.1

Rewrite ${1.06}^{2 t + 2 h}$ as ${({1.06}^{t + h})}^{2}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) (({1.06}^{2 t + 2 h} + {1.06}^{2 t}) ({({1.06}^{t + h})}^{2} - {1.06}^{2 t}))}{h}$

Step 5.4.4.2

Rewrite ${1.06}^{2 t}$ as ${({1.06}^{t})}^{2}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) (({1.06}^{2 t + 2 h} + {1.06}^{2 t}) ({({1.06}^{t + h})}^{2} - {({1.06}^{t})}^{2}))}{h}$

Step 5.4.4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = {1.06}^{t + h}$ and $b = {1.06}^{t}$ .

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) (({1.06}^{2 t + 2 h} + {1.06}^{2 t}) ({1.06}^{t + h} + {1.06}^{t}) ({1.06}^{t + h} - {1.06}^{t}))}{h}$

$\frac{({1.06}^{4 t + 4 h} + {1.06}^{4 t}) ({1.06}^{2 t + 2 h} + {1.06}^{2 t}) ({1.06}^{t + h} + {1.06}^{t}) ({1.06}^{t + h} - {1.06}^{t})}{h}$

Step 6