Practice Exam for Final
First Part
This first batch of questions was mediocre, since we are missing many important parts of the material, and there are many repeated types of problems. Y’all love your population models, apparently. If your question is not in this list, is because there was something else very similar already present. I was expecting to see a few questions on supply and demand, and more on applications to finance.
- The population of the World increased from 4.453 billion in 1980 to 5.397 billion in 1998, and continued at the same percentage rate between 1998 and 2013, and beyond. Express the population as a function of
in years, and use it to compute the projected population in 2015.
- The Chinese population is approximately
with
in billions and
- Find the yearly percent growth rate of the population.
- What was the population in 1994?
- Find the ARC of the population between 1995 and 2005.
- A bank advertises an interest rate of 7.8% per year. If we deposit $6,000 today, how much is in the account 4 years later? If the interest is compounded continuously, how much is in the account 4 years later?
- Find the relative rate of change in the price of a $86 pair of shoes if the price…
- is lowered to $58.50.
- is raised to $105.45$
- The population of Nevada in 2000 was 3.02 million, in 2006, it was 3.598 million. When did the population reach 3.3 million?
- The amount of an antibiotic in the bloodstream decreases by 5.3% a minute. We initially inoculate 125 mg.
- Express the amount of antibiotic in the bloodstream
- When will the amount of antibiotic be 100 mg?
- Express the amount of antibiotic in the bloodstream
- The following table shows the world scooter production in millions between 1950 and 2000.
Year 1950 1960 1970 1980 1990 2000 Scooters 11 20 36 62 92 101 - Find the change in scooter production between 1960 and 1980. Give units.
- Find the average rate of change in the same period of time. Give units and interpret your answer in terms of scooter production.
- Compute
for the function
.
- Determine the
–intercept of the line given by the formula
- A company that makes tables has fixed costs of $15,000 and variable costs of $35 per table. The tables are sold for $140 each.
- Find expressions for the cost and revenue functions.
- Compute the break-even point
Second Part
Whoa! Excellent selection of questions. Fun story problems, pretty much touching every single topic covered in those lectures. Spot on! I even like the formatting of your word document. Great job! I did some small changes and got rid of a couple of repeated questions, or questions badly crafted. Feel free to download from the link below.
Third Part
Another outstanding selection of questions, very thorough, and with lots of nice variations. I did not even mind the repetition of integrals, because they are such an integral part of this part of the course. (lots of puns intended in the previous sentence, of course). There are a few evil questions, which I really appreciate. Kudos to the creator of those (you know who you are)
Fourth Part
I haven’t had time to analyze and process all the questions for part IV. I will do so sometime today. It did look very good at first sight, both in selection and difficulty level.
- Use your calculator to compute the definite integral
.
- The value of a car in 1992 was $25,000. The value of the var decreases as time passes at a continuous rate of 6.4% per year. Assume this trend continues until 2014, when the car is sold. Let
- Write an expression for the value of the car,
, in
- Use left, right and average Riemann sums to approximate the total value of the car between 1992 and 2014 with
and with
.
- Compute the exact total change in the same amount of time
- Write an expression for the value of the car,
- Find the area of the region bounded by the graph of the function
, the
—axis, and the vertical lines
,
.
- A colony of bacteria has a population of 24 million bacteria. Some 4 hours later, the growth rate of said colony is
million bacteria an hour.
- Use an integral to express the change in the amount of bacteria in the first 4 hours.
- What is the population of bacteria at the end of those 4 hours?
- A car travels at a speed of
feet per seconds. How far has the car traveled in the first four minutes?
Francisco Blanco-Silva 
so on question eight in section two, was the word derivation intended to be used? rather than derivative. if so, what does derivation mean?
yup, it is the derivative
Section 2 question 10: are there 4 separate answers or is it supposed to be multiple choice? If multiple choice where are the choices?
Can someone please explain how to do #4 in the first part?
Its relative rate of change the formula used is F(B)-F(A)/F(A) and F(A) for each one is $86 and for F(B) you do it twice and its $58.5 and $105.45
You take the original price, $86 and subtract the new price, $58.50 and divide that all by the original price ($86) and that answer is the percent decrease in price. Same thing for the second part but it’s increase
The formula for relative change is f(b)-f(a)/f(a) use 86 for f(a) and the prices given in each part for f(b)
For section 3 question 11 – what is letter d supposed to be equal to? It just says f(x)=
letter d says f(x) = 1/x^2 +1/x^3 +3√x
can somebody verify that the answer to part 3 number 1 is BELOW
huh?
It is, because marginal profit is greater than marginal cost.
I disagree,
Production=2000, MR exceeds MC, creating MP=.75, since profit is increasing you can expect max profit to occur at a production level higher than 2000 items.
If marginal profit is greater than marginal cost then isn’t the slope at that point positive? If that is true then the profit value is going to increase from that point.
when f(x)= 2000 MR=4 and MC=3.25 therefore MP=4-3.25=.75 if MP is positive then profit will increase from that point
Question 11 section 3, d doesn’t have an answer choice
Those are not answers they are separate questions.
I know that, but letter d was left blank. It’s supposed to read f(x) = 1/x^2 +1/x^3 +3√x
How do you do number 3 on Chapter 3
R=q(p)
R=q (45 – .003q)
R=45q-.003q^2
R = q(p)
R = q (45 – .003q)
R = 45q – .003q squared
In part 3 question 2, is there a typo? Should ln(S/762) be ln(S/763)?
Part 3 question 1….
Production=2000, MR exceeds MC, creating MP=.75, since profit is increasing we can expect max profit to occur at a production level higher than 200 items.
On part 3, question 2, would you set the equation to zero to find out max number of children that could be infected?
you’d first find the first derivative of the original equation, and then set the derivative to zero to find the maximum
What is the final answer for that question?
I got 192 children
? # 2 in 3rd section , how do I set up the equation to solve?
Can anyone explain number 2 on section 3 to me?
It’s a global maximum problem. You take the derivative of the given equation and then set the derivative equal to zero to find the critical points. And then the critical point with the highest value is your answer.
Can you show the steps for this?
45-.003q
45q-.003q^2
Take derivative next so,
45-.006q=0
q=7500
in question four on part four, what would be the starting integral?
would it be the integral 2t+4^t from zero to four?
That’s right. It’d be the integral of 2t+4^t with 4 as the upper limit and 0 as the lower.
Can anyone help me with Exam 4 review #5
what does t represents, (is it seconds, minutes?)
t represents seconds.
how are we supposed to do part 2 question 10
I assume that you just calculate the average between the total and marginal cost
did we ever do anything like that? I’m really confused by this too?
Can someone explain how to do #2 in the fourth part?
Use P=e^(rt) to get the equation and then make a table using the equation you get. You will have to make two tables doing the different changes in time. To find the exact total change you use a definite integral
In number two you are asked to do all of the steps necessary to approximate the total value of the car over the course of 22 years and to calculate the average value of the car for that time as well.
First step (and question) is to create a formula that algebraically describe the value curve for the car.
You are given the initial price and time as well as the slope. You are also told the curve is continuous. So we begin with out basic continuous function (Value)V(t)= Pe^rt. P is given-25,000. e is a constant. r is given by 6.4% per year. t is your independent variable. So
V(t)= 25000e^.064t
The second step is to approximate the total value of the car. Total value is the entire area under the curve bounded by 0 and, in this case, 22. you multiply the step, or x subdivision, by the value y at x sub u. Which is a complicated way to say you multiply (for LRAM) 0 by V(0) plus 11 by V(11) for approximation of the function with a step (delta t) of 2. For RRAM you use the right two values and for average you use the two and average them together. Same steps but with the curve divided into four subdivisions for delta t= 4.
The third step you are asked to calculate the average value of the car from 1982 to 2014. for this you simply evaluate the definite integral of V(t) from 0 to 22.
I didn’t want to just post the answer on here so it required a few more words because I couldn’t just show the work.
My bad I forgot you use the average value theorem for the last part. So just multiply it by 1/b-a
since the value of the car is decreasing, would’t the .064 be negative?
how do you do exam one question 5, I know what i should do I just don’t know what to do with x in the exponent.
If this is your equation: 3.3=3.02(1.0296)^t you divide both sides by 3.02, take the natural log to bring down the t, and divide again.
3.3=3.02(1.0296)^t
(3.3/3.02)=1.0296^t
1.097=1.0296^t
ln(1.097)=ln(1.0296)t
(ln(1.097))/(ln1.0296)=t
3.039=t
how did you get the 1.0296?
You must solve for “a” in the equation. You plug in 6 to the equation 3.02a^6 because it is 6 years from 2000 to 2006. [ P(6)= 3.02a^6 ] This equation is then set equal to 3.598 because it was the population in 2006. So all together you have:
P(6)= 3.02a^6=3.598
a^6=3.598/3.02
a^6=1.19139
Then take the square root of both sides which can be written as a^1/6
a= (1.19139)^1/6
a=1.029
On the first portion of # 2 on part 4 of the exam, would you use P=e^(rt)? and would the rate be negative because the continuous rate is decreasing?
Yes, the formula would be P=25,000e^(-.064×22).
For section 1 #6 I keep getting a – number as my final answer? can anyone help me or is that correct?
For number 6, the equation you write is P=125(.947)^t .
Then you plug in 100 into P to give you 100=125(.947)^t.
Divide both sides by 125 to give .8=(.947)^t
Take the natural log of both sides and bring down the t ln(.8)=ln(.947)t
When you put it in your calculator, be sure to include parenthesis… (ln(.8))/(ln(.947))=t and you will get a positive number
I got t= 4.097 minutes
On part 1, question 10. I’m stuck on finding the equations.
For the cost equation I did: 35q+150000
for the revenue equation i got: 140q
Break even point: Profit equation (140q-15000+35q) set it equal to 0
Make sure you remember to distribute the minus sign in there— 140q-(15000+35q)=0
I ended up with with 142.857 for the break even point
C = 35q + 15000, I think Jenna accidentally typed an extra zero.
In section 1, ? # 4a, is the answer 31.9% ?????
Yes it is
how did you do it?
because I thought f(b) would be the new price: 58.50 and f(a) would be the original, old price: 86
I got -47%; the formula for relative rate of change is f(b)-f(a)/f(a) so it’s 58.50-86/58.5; it’s going to be a negative rate since the price decreases, so once you plug that into your calculator I came out with -.4700 and multiplied x100 to get the percentage for -47%
I think it’s -31.9%. The original f(a) is 86 and f(b) is 58.5. so (58.5-86)/(86)=-.3198, which would be -31.98%
That would make sense. I feel like an idiot for putting the wrong number in the denominator. Thank you!
That is the answer I got
Can someone tell me how they got the rate for the equation for the first question in section one? I feel like I am just confusing myself.
You need to use the equation P=Pa^t.
P=4.453a^t
Then you use the info in the problem to solve for “a”. It is 18 years since 1980 so you plug in P(18), so t=18.
P=4.453a^18
Then you set the equation equal to 5.397 because that was the population in 1998.
5.397=4.453a^18
Then when you solve this you should get a= 1.017
Then you use that value to solve for the population in 2015, where t= 35 because it is 35 years since 1980.
P=4.453(1.017)^35
P=8.033 billion
I used the slope formula y2-y1/x2-x1 and then used that in point slope formula to get my equation! I got 0.052 for my slope/rate
It’s in the notes under exponential functions but you find “a” through the equation: 4.453a^18=5.397 so a^18=1.2119 (which is 5.397/4.453) and to get rid of the exponent to find a you take: a^18*(1/18)=(1.2119)^(1/18), so a=1.0107. I hope this helps!
Does anyone know how to solve question 9 part 2?
I solved it by doing u’v+v’u, which is in the rules of differentiation section of the notes. so u is e^x and v is (3+x)^1/3. the derivative of e^x is e^x then you multiply it by v so you get e^x(3+x)^1/3 then the derivative of (3+x)^1/3 is a chain, so you get 1/3(3+x)^-2/3 e^x, so you’re final answer is: y’= e^x(3+x)^1/3 + 1/3(3+x)^-2/3 e^x
can anyone explain question 13 part 2?
You know that Xo is 3 and you find Yo by doing ln(3) which is 1.099
Then you find slope by finding the derivative of f(x) which is 1/x
Then plug 3 in for x in the derivative to find slope which is 1/3
Then do point slope to get the equation:
y-1.099=1/3(x-3)
y=1/3x=.099
I got the final answer as y=1/3x-2.009… anyone else?
it ends up being y = 1/3x + .098, because 1/3(x-3) comes out to 1/3x – 1. so when you add the -1.098 to both sides, you will end up getting a positive .098.
does that make sense?
I got y=1/3x+.099
I got 1/3x+.099 as well
Test 2, #5: Is the answer 600?
I got $610 because the marginal cost is the derivative of the cost function, which is C= 1500+10q+20q^2 so the derivative is 10+40q so when you plug in 15, you should get 610
I also go $610
I got 610 as well
Can anyone explain how to do section 2 question #17?
N= square root of A
N=sqrt(A)=kA^(1/3)
It’s directly proportional because there is a nonzero constant of K. The graph is increasing and concave down. Larger islands have more species but as the increase slows the island gets larger.
can you explain how you got the second part of that equation? the kA^1/3 part?
I think he meant kA^1/2
Does anyone know how to do section 2 question #20?
Since C(1)=8, and you want to find another value for which C(q)=8, I set the formula equal to 8 and then solved for q using the quadratic equation. This gave me 1 and -3 for values of q, so I believe the answer is -3. However, I’m not positive, so if someone could verify that I’d appreciate it!
I think that you can set the equation to 8… but please if someone else could check me that’d be awesome.
8=q^2 +2q+5
0=q^2 +2q-3
then factor, so 0=(q-1)(q+3)
q= 1 or -3, and the answer is -3.
i could be completely off for this but i thought since it was asking where else the slope is 8 you would set the derivative of that equation equal to 8 since the derivative deals with slope. So i set the derivative equal to 8 and solved, it which case i got q=3.
There was a question like this on test 2 and thats how i solved it for that and i got the answer right so I’m pretty sure thats how you would go about solving it, not plugging it into the original.
For #7 in Part 2, is the graph f(x) or f'(x)
I think its the derivative. C(x) would be (0,5000), not (0,0)
Derivative
For #10 in part 2, what “average” are we supposed to look for?
For #23 in part 2, what exactly is the third part of the question asking, when it says “f ‘ / f”? Is it supposed to be a 2 part question or are you supposed to just use the answer in the first two parts for your “a” and “b”?
That’s right. Use b/a (b is the derivative of a and a is function of 10). So the relative rate of change is 900/10x
I used my answers for the previous questions and solved. So I did 100/900 and got my answer.
Yeah, ignore me, I had it reversed. 100/900 is right.
This question is asking you for the relative rate of change. The relative rate of change of y=f(x) at x=a is f ‘ (a) / f(a), so you are going to use your answers from the first two parts.
How do you find the answer to #7 part 2?
Where ever the slope is the steepest, so in that case at 1.5
7 depicts a graph of a function i assume so to find the greatest derivative you would the single point with the greatest slope. Between like -.6 and .6 the derivative is zero or negative. From there the graph looks identical on each side so I’d say that it is either 1.5 or negative 1.5. I’d probably guess the first. I could be wrong though
I’m not sure whether this is the graph of f(x) or f'(x), but I’ll explain how to find it in both cases..
If it is the graph of f(x), you need to see where the graph is increasing at the steepest slope.
If i is the graph of f'(x), you just need to find the highest point on the graph since the derivative is a graph of the slope. In this case all positive y values indicate a positive slope and all negative values for y indicate a decreasing slope.
For Question 11 on Part 2, do you just plug 3 into the equation?
That’s what I assumed
ok, so I did the average value theorem and I got the same answer as when I just plugged in 3 so I guess either way works?
average value theorem. I posted it below
There is a specific formula used to find the average. I couldn’t figure out a good way to type it so here is a link to a picture of the equation (it is in the baby blue box): http://tutorial.math.lamar.edu/Classes/CalcI/AvgFcnValue.aspx
You’ll use 0 as “a” since at the beginning, it has been 0 years since the start. The value for “b” will be 3. Then you just plug those values into the equation shown in the link.
so once you do all of that, do you subtract 698.716 from 3000 goldfish to find the answer? negative 698.716 is what I got after doing the average value theorem.
How do you do Test 2, Question 11?
Average Value Theorem: 1/(b-a) times integral of f(x) from a to b
f(x)=3000e^(-.4x)
a= 0 b=3
You’ll need to use the average of a function formula, shown in this link (I couldn’t figure out a good way to type it): http://tutorial.math.lamar.edu/Classes/CalcI/AvgFcnValue.aspx
You plug in 0 for “a” since at the start, it has been 0 years. You’ll plug in 3 for “b”. so it will be 1/3-0 times the integral of the equation when you plug in 3 minus the integral of the equation when you plug in zero.
Also has anyone figured out how to do Question 2 on Test 3? I am having a hard time finding the derivative.
First: you need to find the derivative. So step by step for 192ln(S/762) is a u’v+v’u, so u’ = 0, so techincally that part doesn’t matter because you’ll get 0 + v’u, so v’ is ln (S/762) (which is the same thing as ln(a) so the derivative is f(x)/f’(x)) so it’d be 1/762/S/762 which comes out as 1/S because the 762′s cancel each other out. then to finish out this first part you’ll get (1/S)(192) for 192/S. The second part is easier because the derivative of S is just 1, and the derivative of a constant is 0. so the final derivative is: 192/S-1
Second: you have to set the derivative equal to 0 to find the max/mins, so you set 192/S-1=0 and solve. You should get 192=S
Last: check your critical point to see if it is a max/min (hopefully it’ll be a max since that’s the question) and once you do that with the number line test, you should confirm that 192 is the maximum number of children.
I hope this helped; let me know if you’re confused on any part!
Section 3 question 6, could anyone show the steps? I know what to do for integration by parts but I’m not sure if I’m doing the individual parts correctly
sec 2 question 3,,,,,,,,,,derivative of y= 4√x6 ,,,,,,, is the answer 6x^5^1/4 ,,,,, has anyone done that one
Yeah, it helps if you look back at the chain rule functions. f'(x)=g(x)^n then f(x)=g'(x)[ng(x)^(x-1)]After you rewrite it as (x^6)^1/4 you should get the final answer as 6x^5[1/4(x^6)^-3/4]
4√x^6 can also be written like x^((6)(1/4))
f(x)= x^6*1/4, then you can simplify the exponent to be
f(x)=x^(3/2)
And then you can find the derivative of that equation
f’(x)=(3/2)x^(1/2)
Sorry if that’s confusing, it’s kind of difficult to type out, but I hope that helps!
I also got f'(x)=(3/2)x^(1/2)
Can anyone explain what question 10 on part 2 is asking,,, or how to do it?
how do you do question 10 and 14 on section two?
I’m not sure how to solve 10 but 14 is done by adding f ‘(1) to f(1) because f ‘(1) represents the amount it costs to make one more of the item so to find f(2) you add the two together and you should get f(2)=-2.655
4√x^6 can also be written like x^((6)(1/4))
f(x)= x^6*1/4, then you can simplify the exponent to be
f(x)=x^(3/2)
And then you can find the derivative of that equation
f'(x)=(3/2)x^(1/2)
Sorry if that’s confusing, it’s kind of difficult to type out, but I hope that helps!
How do you section 3 question 2? Like the steps of finding the answer
First: you need to find the derivative. So step by step for 192ln(S/762) is a u’v+v’u, so u’ = 0, so techincally that part doesn’t matter because you’ll get 0 + v’u, so v’ is ln (S/762) (which is the same thing as ln(a) so the derivative is f(x)/f'(x)) so it’d be 1/762/S/762 which comes out as 1/S because the 762’s cancel each other out. then to finish out this first part you’ll get (1/S)(192) for 192/S. The second part is easier because the derivative of S is just 1, and the derivative of a constant is 0. so the final derivative is: 192/S-1
Second: you have to set the derivative equal to 0 to find the max/mins, so you set 192/S-1=0 and solve. You should get 192=S
Last: check your critical point to see if it is a max/min (hopefully it’ll be a max since that’s the question) and once you do that with the number line test, you should confirm that 192 is the maximum number of children.
I hope this helped; let me know if you’re confused on any part!
How do you do number 20 on section 3?
integration by parts
Integral(UDV)= U x V – integral(VDU) is integration by parts. Identify U, V, DX and DV and then plug it in.
Right, I understand how to do integration by parts, but I’m messing up at some point and I don’t know where. Can you write out the actual steps? Thanks!
Did you get this?
u=ln(x+2) du= 1/x+2
dv=x^3 v=1/4x^4
isn’t the dv=x^2 and v= 1/3x^3?
Can someone explain section 1 #5 to me??
You must solve for “a” in the equation. You plug in 6 to the equation 3.02a^6 because it is 6 years from 2000 to 2006. [ P(6)= 3.02a^6 ] This equation is then set equal to 3.598 because it was the population in 2006. So all together you have:
P(6)= 3.02a^6=3.598
a^6=3.598/3.02
a^6=1.19139
Then take the square root of both sides which can be written as a^1/6
a= (1.19139)^1/6
a=1.029
Then solve for t in this equation: 3.3=3.02(1.029)^t by dividing and then taking the “ln” of both sides.
You use the formula: P=Poa^t for the information you’re given to find the rate which the population grows and then you plug it back in.
Here’s how I did it:
P=Poa^t
3.598=3.02a^6
(3.598/3.02)^(1/6)=a
1.03=a
Then you plug this into the other equation to find the time:
3.3=3.02(1.03)^t
ln(3.3/3.02)=ln1.03^t
t=2.9
Therefore, at the growth rate we found, it takes about 3 years for the population of Nevada to grow to 3.3 million people.
5.The population of Nevada in 2000 was 3.02 million, in 2006, it was 3.598 million. When did the population reach 3.3 million?
First you need to find the equation, P=P(a)^t so the P is the original population, 3.02. The “a” is the rate and it is what you need to find. Then the t is the amount of time between 2006 and 2000, so 6. Then to find “a” you set the equation 3.02(a)^6 equal to 3.598 because that is the population in 2006. Then solve for “a”. You should get 1.0296
Next to find when the population reached 3.3 million you set the equation… 3.02(1.0296)^t=3.3 and solve for t. At that point you should get 3.03 for t and that would make the answer 2003.
If you have any other trouble we did a problem similar to this one in class at the beginning of the year, I believe it was lesson 5..hope that helps!
3.598=3.02e^r*t where t= 6 is as far as I got with that one ,,,, need “r” to solve
Can someone help me with section 2 number 4 and 9? They are derivative problems I just forgot how to do them.
On #4 the derivative of an ln is f'(x)/f(x).
On #9 you have to use the product rule, which is the derivate of the first term times the second term plus the derivative of the second term times the first term.
Does this make sense?
an easier way to say for #9 is: u’v+v’u
4. 20t^4 / 4t^5+3
9. (e^x * 1/3(3+x)^-2/3) + e^x(3+x)^(1/3)
For Part 4 Question 3 is the equation supposed to be x^3-8x^2+15x+1?
I think there is a typo in the problem because it says 15+1….
I think it is a typo. Just try solving it both ways.
Test 3, Question 4: I am getting x= -3/2 when taking the derivative and solving for the critical points. But the interval is when x>0 and -3/2 is less than 0. Does anyone know what I’m doing wrong?
I’m having the same problem..
Yeah I did the same thing. I ended up just graphing it on my calculator to figure out the min/max and I got (1,5) for the minimum and infinity for the max because -3/2 didn’t fit into the interval
For part two, question ten, are we just supposed to find the average?
I assumed so… it looks like it was multiple choice and the answers got deleted
Also, the cheat sheets will be attached to the exam right? I think I remember him saying that in class but not 100% sure. If anyone knows for sure let me know please. Thanks!
I’m pretty sure that we are going to have it for the exam!
yeah that makes sense thank you!
Hey y’all! Is there a section in our notes that would help me answer #7 of section 2? I’m pretty sure I have the answer but I just want to make sure I’m doing it correctly and I can’t find it in my notes. Thank you!
There’s an example like #7 in Lesson 7 of our notes. Hope that helps!
Thank you!!!
Question #13 on Section 13, once you find the first derivative and set it equal to 0, what do you do? You get f(x) = 5x^4-15x^2+5 which you can’t put in the quadratic formula.
section 3, excuse me.
You can graph it on your calculator and find the zeros that way.
sooo…. wheres the rest of part four?
Which questions in section 3 were marked as “evil”?
Numbers 11 and 12
can anyone explain number 16 on test number 3?
can anyone explain question 16 on test 3?
Revenue = price * quantity;
p = 46 – 0.04q;
R = (46-.04q)q
R = 46q – .04q^2
Revenue= price x quantity
So the revenue function is R= (46-.04q) (q)
Then to find the price and quantity that maximizes profit, take the derivative and set it equal to 0.
R= (46q-.04q^2)=0
you should get q= 575 units. the price per unit is $23. The revenue at that level is then $13225
Can someone please explain question 4 of section 3. I understand you use the first derivative, equal it to zero and get a point of (-3/2,-9/4) but how do you know if thats a max or a min? And how do you find the other global point?
Also, this isnt a pat of the practice sections but it was on one of our tests:
Find a constant a so that f(x)_x^2+ax+5 has a local minimum at 3,4
Take the derivate of f(x) which gives you 2x+a. Plug in 3 for x and set it equal to 0 and solve for a.
2(3)+a=0
a=-6
One more question….can someone explain how to do number 13 in section 2? I know this is pretty basic but I’m having a block.
you want to use the point slope equation, y-y1=m(x=x1)
x1 you use 3
you plug 3 into the function for y1 and you get ln(3) which is 1.0986
and then find the derivative of the function which the derivative a ln(x) is 1/x, and you plus 3 into that to find the slope, m, which is 1/3.
plug all of those into the equation and get it into y-intercept form,
you end with y= 1/3x + .09 if i did the simplification right i believe
typo: the equation is y-y1=m(x-x1)
How do I do section 4 number 2??
P=Pe^rt is the formula you need to use to find the expression. So you get P=25,000e^-.064t, with t being the years since 1992.
Do you need help with part 2 and 3 of the question too?
Can someone please explain #7 on section 2?
The derivative is essentially the slope of a point on a line that is tangent to a the curve. So the derivative would be the largest when the rise:run ratio is both positive and the largest. The .gif on this page (http://en.wikipedia.org/wiki/Derivative) explains it visually.
Green is positive and red is negative. When the green tangent line is both positive and the steepest, that is when the derivative is the largest.
There’s an example in our notes in Lesson 7 that might help you understand it. You are just looking at the graph and determining where the slope is the steepest. Also review Test 2 for more examples. Hope this helps!
I am not sure if the graph is the graph of the function or the graph of the derivative but heres how you do it for both cases:
if its the graph of the function then you want to find where the stope is the steepest on the graph because then the derivative would be largest
if its the graph of the derivative you want to look for the highest point on the graph because thats where the derivative would be the largest as well
hope that helps!
For question 13 on section 3, simply find the first derivative of 5x^4-15x^2 +5 and set it equal to zero then solve. With that x should equal the square root of 3. Then continue by finding the second derivative, by solving x should equal 0.
Could someone please help me with understanding question 2 in section 3 please.
Anyone know how to do Question 17 on Section 3?
Profit is revenue – cost so for this one you can get the equation:
P=487q-(30,000+4q^2)
To find when profit is maximized you need to take the derivative and set it equal to 0.
When I did this I got that profit is maximized when q=60.875
Then, the find what the profit would be at this point, plug 60.875 back into your original profit equation!
Profit maximization is where MR = MC. So find the derivative of the cost function and set it equal to the derivative of the revenue function and solve to find the answer.
section 3 question 6.. which one did you use as u and which as dv?
u should be (2x-3)^1/2 and dv should be e^-2x
Anything with ” e”is “Dv”
Anything with “ln” is “U”
That’s wat I have in my notes.
for question 6 i used e^-2x as my u and (2x-3)^1/2 as my dv
but yeah i agree with zach, for this problem i don’t think it really matters what you pick, just whatever you think you could easier take the anti derivative and derivative of , the only ones i think it matters for is when you have ln(x)
Can anyone show me the steps how to do last question on Part 1?
Cost function is just the fixed cost plus variable costs, so the fixed cost is $15,000 and the variable cost is $35 per table, so the cost function is: C(x)= 35x+15,000
Revenue function is how much the tables will sell for, so $140 per table, so the revenue function is R(x)= 140x
The break even point is when cost and revenue equal each other, so 140x=35x+15,000 and when you solve that you should get x= 142.85, or 143 tables.
The cost function would be the fixed cost plus the variable costs, so C= 35x + 15,000. The revenue function is R= 140x. To find the break even point all you do is set the two functions equal to each other and solve for x.
So 35x+15,000 = 140x
15,000 = 105x
142.85 = x
for the cost and revenue functions you just plug in the values but to find the break even point you can either set the profit equation = 0 or you can set equal the cost and revenue functions!