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Precalculus : Introductory Calculus

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Example Questions

Example Question #2 : Rate Of Change Problems

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 #4 : 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:

$-\$5.39$

$\$38.73$

$\$5.39$

$\$38.39$

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 #5 : 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 #4 : 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 #5 : 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 #6 : 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 horizontal on that interval.

We know f(x) is a polynomial.

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

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

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

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 : Derivatives

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 odd

f(x) is increasing

f(x) is decreasing

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 decreasing

f(x) is negative

f(x) is an odd function

f(x) is constant

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 #1 : Maximum And Minimum Problems

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:

$\small 42000$

$\small 3000$

$\small 21000$

$\small 7$

$\small 7000$

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 #2 : Maximum And Minimum Problems

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

Possible Answers:

Critical Point=(0,2)

Minimum

Critical Point=(0,2)

Max

Critical Point=(0,0)

Minimum

Critical Point=(2,0)

Minimum

Critical Point=(0,0)

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 #2 : Rate Of Change Problems

Example Question #4 : Rate Of Change Problems

Example Question #5 : Rate Of Change Problems

Example Question #4 : Rate Of Change Problems

Example Question #5 : Rate Of Change Problems

Example Question #6 : Rate Of Change Problems

Example Question #11 : Derivatives

Example Question #12 : Rate Of Change Problems

Example Question #1 : Maximum And Minimum Problems

Example Question #2 : Maximum And Minimum Problems

All Precalculus Resources

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