Precalculus Examples

$f (t) = 21 t - 3 t^{2}$ ; $0 \leq t \leq 9$

Step 1

Write $f (t) = 21 t - 3 t^{2}$ as an equation.

$y = 21 t - 3 t^{2}$

Step 2

Substitute using the average rate of change formula.

Tap for more steps...

Step 2.1

The average rate of change of a function can be found by calculating the change in $y$ values of the two points divided by the change in $t$ values of the two points.

$\frac{f (9) - f (0)}{(9) - (0)}$

Step 2.2

Substitute the equation $y = 21 t - 3 t^{2}$ for $f (9)$ and $f (0)$ , replacing $t$ in the function with the corresponding $t$ value.

$\frac{(21 (9) - 3 {(9)}^{2}) - (21 (0) - 3 {(0)}^{2})}{(9) - (0)}$

Step 3

Simplify the expression.

Tap for more steps...

Step 3.1

Simplify the numerator.

Tap for more steps...

Step 3.1.1

Multiply $21$ by $9$ .

$\frac{189 - 3 \cdot 9^{2} - (21 (0) - 3 \cdot 0^{2})}{9 - (0)}$

Step 3.1.2

Raise $9$ to the power of $2$ .

$\frac{189 - 3 \cdot 81 - (21 (0) - 3 \cdot 0^{2})}{9 - (0)}$

Step 3.1.3

Multiply $- 3$ by $81$ .

$\frac{189 - 243 - (21 (0) - 3 \cdot 0^{2})}{9 - (0)}$

Step 3.1.4

Simplify each term.

Tap for more steps...

Step 3.1.4.1

Multiply $21$ by $0$ .

$\frac{189 - 243 - (0 - 3 \cdot 0^{2})}{9 - (0)}$

Step 3.1.4.2

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

$\frac{189 - 243 - (0 - 3 \cdot 0)}{9 - (0)}$

Step 3.1.4.3

Multiply $- 3$ by $0$ .

$\frac{189 - 243 - (0 + 0)}{9 - (0)}$

Step 3.1.5

Add $0$ and $0$ .

$\frac{189 - 243 - 0}{9 - (0)}$

Step 3.1.6

Multiply $- 1$ by $0$ .

$\frac{189 - 243 + 0}{9 - (0)}$

Step 3.1.7

Add $189 - 243$ and $0$ .

$\frac{189 - 243}{9 - (0)}$

Step 3.1.8

Subtract $243$ from $189$ .

$\frac{- 54}{9 - (0)}$

Step 3.2

Simplify the denominator.

Tap for more steps...

Step 3.2.1

Multiply $- 1$ by $0$ .

$\frac{- 54}{9 + 0}$

Step 3.2.2

Add $9$ and $0$ .

$\frac{- 54}{9}$

Step 3.3

Divide $- 54$ by $9$ .

$- 6$