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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1 : Derivatives

Suppose we can model the profit, , in dollars from selling items with the equation P(x)=3x^2-4x+2 .

Find the average rate of change of the profit from x=1 to x=4 .

Possible Answers:

Correct answer:

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 P(4)=3(4)^2-4(4)+2=3(16)-16+2=48-16+2=34 , and P(1)=3(1)^2-4(1)+2=3(1)-4+2=3-4+2=1 , we find that the average rate of change is $\frac{34-1}{4-1}=\frac{33}{3}=11$ .

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Example Question #2 : Derivatives

Let the profit, , (in thousands of dollars) earned from producing items be found by P(x)=2x^2-6x+25 .

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

Possible Answers:

$\frac{4}{3}$

Correct answer:

Explanation:

Since P(5)=2(5)^2-6(5)+25 , we see that this equals 2(25)-30+25=50-30+25=20+25=45 . Now let's examine P(4) . P(4)=2(4)^2-6(4)+25=2(16)-24+25 which simplifies to .

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

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Example Question #5 : Derivatives

Suppose that a customer purchases dog treats based on the sale price , where D(p)=15-p^2+p , 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.

Possible Answers:

Correct answer:

Explanation:

D(3)=15-9+3=6+3=9

D(2)=15-4+2=13

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.

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Example Question #6 : Derivatives

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 t=0 and t=3 , the year the student expects to graduate.

Possible Answers:

$-\$5.39$

$\$38.73$

$\$38.39$

$\$5.39$

Correct answer:

$\$5.39$

Explanation:

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

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

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

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Example Question #1 : Rate Of Change Problems

Find the average rate of change of $y=\sqrt{x-3}$ between x=4 and x=7 .

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

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

Here f(7) 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 $\frac{2-1}{7-4}=\frac{1}{3}$ .

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Example Question #1 : Rate Of Change Problems

Find the average rate of change of y=2x^2+3x over the interval from x=1 to x=3 .

Possible Answers:

Correct answer:

Explanation:

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

f(3)=2(3)^2+3(3)=2(9)+9=18+9=27 .

f(1)=2(1)^2+3(1)=2(1)+3=2+3=5 .

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

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Example Question #2 : Rate Of Change Problems

Find the average rate of change of y=-x^2+3x-6 over the interval from x=-2 to x=0 .

Possible Answers:

Correct answer:

Explanation:

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

Now f(0)=-(0)^2+3(0)-6=-0+0-6=-6 .

We also know f(-2)=-(-2)^2+3(-2)-6=-4-6-6=-16 .

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

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Example Question #2 : Rate Of Change Problems

Why can we make an educated guess that the average rate of change of f(x)=x^2 , between x=-1 and x=1 would be ?

Possible Answers:

We know f(x) is vertical on that interval.

We know f(x) is symmetrical on that interval.

We know f(x) is horizontal on that interval.

We know f(x) is odd on that interval.

We know f(x) is a polynomial.

Correct answer:

We know f(x) is symmetrical on that interval.

Explanation:

Because f(x) 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 .

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Example Question #11 : Rate Of Change Problems

If the average rate of change of f(x) between x=a and x=b , where b>a , is positive, then what can be said about f(x) on that interval?

Possible Answers:

f(x) is odd

f(x) is decreasing

f(x) is constant

f(x) is increasing

Correct answer:

f(x) is increasing

Explanation:

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

f(b)-f(a)>0 and f(b)>f(a) . This shows that f(x) must increase over the interval from to .

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Example Question #12 : Rate Of Change Problems

If the average rate of change of f(x) between x=a and x=b , where a<b , is negative, then what can be said about f(x) on that interval?

Possible Answers:

f(x) is constant

f(x) is an odd function

f(x) is negative

f(x) is decreasing

f(x) is increasing

Correct answer:

f(x) is decreasing

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.

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

f(b)-f(a)<0

f(b)<f(a) and since a<b , f(x) is decreasing

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #5 : Derivatives

Example Question #6 : Derivatives

Example Question #1 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

Example Question #2 : Rate Of Change Problems

Example Question #2 : Rate Of Change Problems

Example Question #11 : Rate Of Change Problems

Example Question #12 : Rate Of Change Problems

All Precalculus Resources

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