Pre-Calculus - Rate of Change Problems

Let a function be defined by .

Find the average rate of change of the function over .

$-\frac{ln5}{4}$

$-ln\frac{5}{4}$

$ln\frac{-5}{4}$

$\frac{ln5}{4}$

Explanation

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 .

Explanation

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 $\frac{34-1}{4-1}=\frac{33}{3}=11$ .

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

is increasing

is decreasing

is constant

is odd

Explanation

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?

is decreasing

is negative

is an odd function

is constant

is increasing

Explanation

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

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.

$\frac{4}{3}$

Explanation

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.

Explanation

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.

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

Explanation

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 over the interval from to .

Explanation

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

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

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

We know is symmetrical on that interval.

We know is horizontal on that interval.

We know is vertical on that interval.

We know is odd on that interval.

We know is a polynomial.

Explanation

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 .