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Precalculus : Introductory Calculus

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Example Questions

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 #1 : Rate Of Change Problems

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:

$-\$5.39$

$\$5.39$

$\$38.73$

$\$38.39$

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 #1 : 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 #51 : Introductory Calculus

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 constant

f(x) is decreasing

f(x) is increasing

f(x) is odd

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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Example Question #51 : Introductory Calculus

The profit of a certain cellphone manufacturer can be represented by the function

$\small \small p(x)=-2000x^3+12000x^2+126000x$

where $\small p$ is the profit in dollars and $\small x$ is the production level in thousands of units. How many units should be produced to maximize profit?

Possible Answers:

$\small 3000$

$\small 21000$

$\small 42000$

$\small 7000$

$\small 7$

Correct answer:

$\small 7000$

Explanation:

We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.

$\small p'(x)=3(-2000x^2)+2(12000x)+126000=-6000x^2+24000x+126000$

We then set this derivative equal to 0 and solve.

$\small 0=-6000x^2+24000x+126000$

We can factor out our greatest common factor and then divide it out.

$\small 0=-6000(x^2-4x-21)$

$\small 0=x^2-4x-21$

Next we can factor.

$\small 0=(x+3)(x-7)$

Therefore, we can solve

$\small x=-3\:or\:7$

We remember that these production levels are in thousands of units.

However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.

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Example Question #1 : Maximum And Minimum Problems

What is the critical point of y=x^2 . Is it a max or a minimum?

Possible Answers:

Critical Point=(0,0)

Minimum

Critical Point=(0,0)

Max

Critical Point=(2,0)

Minimum

Critical Point=(0,2)

Minimum

Critical Point=(0,2)

Max

Correct answer:

Critical Point=(0,0)

Minimum

Explanation:

To find the critical points, we first take the first derivative using the Power Rule:

$y=x^n \rightarrow y'=nx^{n-1}$

Therefore,

y=x^2

y'=2x .

To find the critical point we need to set the derivative equal to zero and solve for x.

Doing so this gives the critical point at x=0 . To get the y value of the point plug x=0 back into the original equation.

y=0^2=0 thus the point is (0,0) .

Is it a max or min?

Take the second derivative

y''=2

It is a minimum since the second derivative is positive.

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Precalculus : Introductory Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #5 : Derivatives

Example Question #1 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

Example Question #2 : Rate Of Change Problems

Example Question #51 : Introductory Calculus

Example Question #12 : Rate Of Change Problems

Example Question #51 : Introductory Calculus

Example Question #1 : Maximum And Minimum Problems

All Precalculus Resources

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