Rate of Change Problems - Pre-Calculus

Card 0 of 48

Find the average rate of change of over the interval from to .

Tap to see back →

The average rate of change will be $\frac{f(3)-f(1)}{3-1}$ .

.

.

This gives us $$\frac{27-5}{3-1}$=\frac{22}{2}$=11$ .

Find the average rate of change of over the interval from to .

Tap to see back →

The average rate of change will be $\frac{f(0)-f(-2)}{0-(-2)}$ .

Now.

We also know .

So we have $$\frac{-6-(-16)}{0-(-2)}$=\frac{-6+16}{0+2}$=\frac{10}{2}$=5$ .

Find the average rate of change of $y=$\sqrt{x-3}$$ between and .

Tap to see back →

The solution will be found by the formula $\frac{f(7)-f(4)}{7-4}$ .

Here gives us $$\sqrt{7-3}$=$\sqrt{4}$=2$ , and $f(4)=$\sqrt{4-3}$=$\sqrt{1}$=1$ .

Thus, we find that the average rate of change is $\frac{2-1}{7-4}$=\frac{1}{3}$ .

Find the average rate of change of the function over the interval from to .

Tap to see back →

The average rate of change will be found by $\frac{f(2)-f(0)}{2-0}$ .

Here, , and .

Now, we have .

Let a function be defined by .

Find the average rate of change of the function over .

Tap to see back →

We use the average rate of change formula, which gives us $\frac{f(4)-f(0)}{4-0}$ .

Now , and .

Therefore, the answer becomes $\frac{-ln5-0}{4-0}$=-$\frac{ln5}{4}$ .

Suppose we can model the profit, , in dollars from selling items with the equation .

Find the average rate of change of the profit from to .

Tap to see back →

We need to apply the formula for the average rate of change to our profit equation. Thus we find the average rate of change is $\frac{P(4)-P(1)}{4-1}$ .

Since , and , we find that the average rate of change is $$\frac{34-1}{4-1}$=\frac{33}{3}$=11$ .

Let the profit, , (in thousands of dollars) earned from producing items be found by .

Find the average rate of change in profit when production increases from 4 items to 5 items.

Tap to see back →

Since , we see that this equals. Now let's examine . which simplifies to .

Therefore the average rate of change formula gives us $$\frac{f(5)-f(4)}{5-4}$=\frac{45-33}{5-4}$=\frac{12}{1}$=12$ .

Suppose that a customer purchases dog treats based on the sale price , where , where $1\leq p\leq3$ .

Find the average rate of change in demand when the price increases from $2 per treat to $3 per treat.

Tap to see back →

Thus the average rate of change formula yields $$\frac{9-13}{3-2}$=\frac{-4}{1}$=-4$ .

This implies that the demand drops as the price increases.

A college freshman invests $100 in a savings account that pays 5% interest compounded continuously. Thus, the amount saved after years can be calculated by $A(t)=100e^{0.05t}$ .

Find the average rate of change of the amount in the account between and , the year the student expects to graduate.

Tap to see back →

$A(3)=100e^{0.05*3}=100e^{0.15}\approx \$116.18$ .

$A(0)=100e^{0.05\ast0}=100e^{0}=\$100.00$ .

Hence, the average rate of change formula gives us $\frac{A(3)-A(0)}{3-0}=\frac{$116.18-$100.00}{3-0}=\frac{$16.18}{3}\approx $5.39$ .

Why can we make an educated guess that the average rate of change of , between and would be ?

Tap to see back →

Because is symmetrical over the y axis, it increases exactly as much as it decreases on the interval from to . Thus the average rate of change on that interval will be .

If the average rate of change of between and , where , is positive, then what can be said about on that interval?

Tap to see back →

If the average rate of change is positive, then the formula gives us $$\frac{f(b)-f(a)}{b-a}$>0$ , so . We know because it is a given in the proble, so . Hence

and . This shows that must increase over the interval from to .

If the average rate of change of between and , where , is negative, then what can be said about on that interval?

Tap to see back →

If the average rate of change is negative, then the function is changing in a negative direction overall. Hence, the graph of the function will be decreasing on that interval.

$$\frac{f(b)-f(a)}{b-a}$<0$

and since , is decreasing

Find the average rate of change of over the interval from to .

Tap to see back →

The average rate of change will be $\frac{f(3)-f(1)}{3-1}$ .

.

.

This gives us $$\frac{27-5}{3-1}$=\frac{22}{2}$=11$ .

Find the average rate of change of over the interval from to .

Tap to see back →

The average rate of change will be $\frac{f(0)-f(-2)}{0-(-2)}$ .

Now.

We also know .

So we have $$\frac{-6-(-16)}{0-(-2)}$=\frac{-6+16}{0+2}$=\frac{10}{2}$=5$ .

Find the average rate of change of $y=$\sqrt{x-3}$$ between and .

Tap to see back →

The solution will be found by the formula $\frac{f(7)-f(4)}{7-4}$ .

Here gives us $$\sqrt{7-3}$=$\sqrt{4}$=2$ , and $f(4)=$\sqrt{4-3}$=$\sqrt{1}$=1$ .

Thus, we find that the average rate of change is $\frac{2-1}{7-4}$=\frac{1}{3}$ .

Find the average rate of change of the function over the interval from to .

Tap to see back →

The average rate of change will be found by $\frac{f(2)-f(0)}{2-0}$ .

Here, , and .

Now, we have .

Let a function be defined by .

Find the average rate of change of the function over .

Tap to see back →

We use the average rate of change formula, which gives us $\frac{f(4)-f(0)}{4-0}$ .

Now , and .

Therefore, the answer becomes $\frac{-ln5-0}{4-0}$=-$\frac{ln5}{4}$ .

Suppose we can model the profit, , in dollars from selling items with the equation .

Find the average rate of change of the profit from to .

Tap to see back →

We need to apply the formula for the average rate of change to our profit equation. Thus we find the average rate of change is $\frac{P(4)-P(1)}{4-1}$ .

Since , and , we find that the average rate of change is $$\frac{34-1}{4-1}$=\frac{33}{3}$=11$ .

Let the profit, , (in thousands of dollars) earned from producing items be found by .

Find the average rate of change in profit when production increases from 4 items to 5 items.

Tap to see back →

Since , we see that this equals. Now let's examine . which simplifies to .

Therefore the average rate of change formula gives us $$\frac{f(5)-f(4)}{5-4}$=\frac{45-33}{5-4}$=\frac{12}{1}$=12$ .

Suppose that a customer purchases dog treats based on the sale price , where , where $1\leq p\leq3$ .

Find the average rate of change in demand when the price increases from $2 per treat to $3 per treat.

Tap to see back →

Thus the average rate of change formula yields $$\frac{9-13}{3-2}$=\frac{-4}{1}$=-4$ .

This implies that the demand drops as the price increases.