Precalculus Examples

$d (t) = 0.8 t^{2}$

Step 1

Consider the difference quotient formula.

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

Step 2

Find the components of the definition.

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Step 2.1

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

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Step 2.1.1

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

$d (t + h) = 0.8 {(t + h)}^{2}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

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

$d (t + h) = 0.8 ((t + h) (t + h))$

Step 2.1.2.2

Expand $(t + h) (t + h)$ using the FOIL Method.

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Step 2.1.2.2.1

Apply the distributive property.

$d (t + h) = 0.8 (t (t + h) + h (t + h))$

Step 2.1.2.2.2

Apply the distributive property.

$d (t + h) = 0.8 (t \cdot t + t h + h (t + h))$

Step 2.1.2.2.3

Apply the distributive property.

$d (t + h) = 0.8 (t \cdot t + t h + h t + h \cdot h)$

Step 2.1.2.3

Simplify and combine like terms.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Multiply $t$ by $t$ .

$d (t + h) = 0.8 (t^{2} + t h + h t + h \cdot h)$

Step 2.1.2.3.1.2

Multiply $h$ by $h$ .

$d (t + h) = 0.8 (t^{2} + t h + h t + h^{2})$

Step 2.1.2.3.2

Add $t h$ and $h t$ .

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Step 2.1.2.3.2.1

Reorder $t$ and $h$ .

$d (t + h) = 0.8 (t^{2} + h t + h t + h^{2})$

Step 2.1.2.3.2.2

Add $h t$ and $h t$ .

$d (t + h) = 0.8 (t^{2} + 2 h t + h^{2})$

Step 2.1.2.4

Apply the distributive property.

$d (t + h) = 0.8 t^{2} + 0.8 (2 h t) + 0.8 h^{2}$

Step 2.1.2.5

Multiply $2$ by $0.8$ .

$d (t + h) = 0.8 t^{2} + 1.6 h t + 0.8 h^{2}$

Step 2.1.2.6

The final answer is $0.8 t^{2} + 1.6 h t + 0.8 h^{2}$ .

$0.8 t^{2} + 1.6 h t + 0.8 h^{2}$

Step 2.2

Reorder.

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Step 2.2.1

Move $0.8 t^{2}$ .

$1.6 h t + 0.8 h^{2} + 0.8 t^{2}$

Step 2.2.2

Reorder $1.6 h t$ and $0.8 h^{2}$ .

$0.8 h^{2} + 1.6 h t + 0.8 t^{2}$

Step 2.3

Find the components of the definition.

$d (t + h) = 0.8 h^{2} + 1.6 h t + 0.8 t^{2}$

$d (t) = 0.8 t^{2}$

$d (t + h) = 0.8 h^{2} + 1.6 h t + 0.8 t^{2}$

$d (t) = 0.8 t^{2}$

Step 3

Plug in the components.

$\frac{d (t + h) - d t}{h} = \frac{0.8 h^{2} + 1.6 h t + 0.8 t^{2} - (0.8 t^{2})}{h}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Factor $0.2$ out of $0.8 h^{2} + 1.6 h t + 0.8 t^{2} - 1 \cdot 0.8 t^{2}$ .

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Step 4.1.1.1

Move $h$ .

$\frac{0.8 h^{2} + 1.6 t h + 0.8 t^{2} - 1 \cdot 0.8 t^{2}}{h}$

Step 4.1.1.2

Factor $0.2$ out of $0.8 h^{2}$ .

$\frac{0.2 (4 h^{2}) + 1.6 t h + 0.8 t^{2} - 1 \cdot 0.8 t^{2}}{h}$

Step 4.1.1.3

Factor $0.2$ out of $1.6 t h$ .

$\frac{0.2 (4 h^{2}) + 0.2 (8 t h) + 0.8 t^{2} - 1 \cdot 0.8 t^{2}}{h}$

Step 4.1.1.4

Factor $0.2$ out of $0.8 t^{2}$ .

$\frac{0.2 (4 h^{2}) + 0.2 (8 t h) + 0.2 (4 t^{2}) - 1 \cdot 0.8 t^{2}}{h}$

Step 4.1.1.5

Factor $0.2$ out of $- 1 \cdot 0.8 t^{2}$ .

$\frac{0.2 (4 h^{2}) + 0.2 (8 t h) + 0.2 (4 t^{2}) + 0.2 (- 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.1.6

Factor $0.2$ out of $0.2 (4 h^{2}) + 0.2 (8 t h)$ .

$\frac{0.2 (4 h^{2} + 8 t h) + 0.2 (4 t^{2}) + 0.2 (- 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.1.7

Factor $0.2$ out of $0.2 (4 h^{2} + 8 t h) + 0.2 (4 t^{2})$ .

$\frac{0.2 (4 h^{2} + 8 t h + 4 t^{2}) + 0.2 (- 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.1.8

Factor $0.2$ out of $0.2 (4 h^{2} + 8 t h + 4 t^{2}) + 0.2 (- 5 \cdot 0.8 t^{2})$ .

$\frac{0.2 (4 h^{2} + 8 t h + 4 t^{2} - 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.2

Factor using the perfect square rule.

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Step 4.1.2.1

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

$\frac{0.2 ({(2 h)}^{2} + 8 t h + 4 t^{2} - 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.2.2

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

$\frac{0.2 ({(2 h)}^{2} + 8 t h + {(2 t)}^{2} - 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 t h = 2 \cdot (2 h) \cdot (2 t)$

Step 4.1.2.4

Rewrite the polynomial.

$\frac{0.2 ({(2 h)}^{2} + 2 \cdot (2 h) \cdot (2 t) + {(2 t)}^{2} - 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.2.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 h$ and $b = 2 t$ .

$\frac{0.2 ({(2 h + 2 t)}^{2} - 5 \cdot 0.8 t^{2})}{h}$

Step 4.1.3

Rewrite $5 \cdot 0.8 t^{2}$ as ${(2 t)}^{2}$ .

$\frac{0.2 ({(2 h + 2 t)}^{2} - {(2 t)}^{2})}{h}$

Step 4.1.4

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

$\frac{0.2 ((2 h + 2 t + 2 t) (2 h + 2 t - (2 t)))}{h}$

Step 4.1.5

Simplify.

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Step 4.1.5.1

Add $2 t$ and $2 t$ .

$\frac{0.2 ((2 h + 4 t) (2 h + 2 t - (2 t)))}{h}$

Step 4.1.5.2

Factor $2$ out of $2 h + 4 t$ .

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Step 4.1.5.2.1

Factor $2$ out of $2 h$ .

$\frac{0.2 ((2 h + 4 t) (2 h + 2 t - (2 t)))}{h}$

Step 4.1.5.2.2

Factor $2$ out of $4 t$ .

$\frac{0.2 ((2 h + 2 (2 t)) (2 h + 2 t - (2 t)))}{h}$

Step 4.1.5.2.3

Factor $2$ out of $2 h + 2 (2 t)$ .

$\frac{0.2 (2 (h + 2 t) (2 h + 2 t - (2 t)))}{h}$

Step 4.1.5.3

Multiply $2$ by $- 1$ .

$\frac{0.2 (2 (h + 2 t) (2 h + 2 t - 2 t))}{h}$

$\frac{0.2 \cdot 2 (h + 2 t) (2 h + 2 t - 2 t)}{h}$

Step 4.1.6

Multiply $0.2$ by $2$ .

$\frac{0.4 (h + 2 t) (2 h + 2 t - 2 t)}{h}$

Step 4.1.7

Subtract $2 t$ from $2 t$ .

$\frac{0.4 (h + 2 t) (2 h + 0 t)}{h}$

Step 4.1.8

Multiply $0$ by $t$ .

$\frac{0.4 (h + 2 t) (2 h + 0)}{h}$

Step 4.1.9

Add $2 h$ and $0$ .

$\frac{0.4 (h + 2 t) \cdot 2 h}{h}$

Step 4.1.10

Multiply $2$ by $0.4$ .

$\frac{0.8 (h + 2 t) h}{h}$

Step 4.2

Simplify terms.

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Step 4.2.1

Cancel the common factor of $h$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{0.8 (h + 2 t) h}{h}$

Step 4.2.1.2

Divide $0.8 (h + 2 t)$ by $1$ .

$0.8 (h + 2 t)$

Step 4.2.2

Apply the distributive property.

$0.8 h + 0.8 (2 t)$

Step 4.2.3

Simplify the expression.

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Step 4.2.3.1

Multiply $2$ by $0.8$ .

$0.8 h + 1.6 t$

Step 4.2.3.2

Reorder $0.8 h$ and $1.6 t$ .

$1.6 t + 0.8 h$

Step 5