Review for Final Exam—Part 1
- Find the horizontal and vertical asymptotes of the function
- Compute
- At which values of
is the function
discontinuous?
- Use the definition of derivative as a limit to compute the derivative of
- There are exactly two lines that are tangent to the curve
and parallel to the line
Find their equations.
- Compute the derivative of the function
- Find the tangent line to the curve
at
- Find the slope of the tangent line to the curve defined by
at the point
- A cylindrical water tank of radius 5m and length 20m which is lying on its side (i.e., its circular ends are vertical) is being filled at the rate of 100m3/s. How quickly is the surface rising when the water is 7m deep?
- Find the absolute minimum and absolute maximum of the function
on the interval
- Let
- Find the intervals on which
is increasing or decreasing.
- Find all local minimum and maximum values of
- Find the intervals of convexity and the inflection points of
- Find the intervals on which
- Compute
- Approximate the area under the curve
between
and
by taking a Riemann sum with 4 regions, using the right-endpoint rule.
- Compute the indefinite integral:
![[1/2, 3].](https://s0.wp.com/latex.php?latex=%5B1%2F2%2C+3%5D.&bg=ffffff&fg=555555&s=0&c=20201002)
Francisco Blanco-Silva 
Hey does anyone where to start question 3? In recitation we did one were there was 3 equations and I know that in a system of equation, the points where there is a possible problem with continuity is “where the line breaks up”. for example, in this problem the possible x-values where the system of equation could be discontinuous would be at x=-1,1,3. So the starting information that I know is that we look at these values as the limit approach from the left and right of each x-value. But I dont know where to go from there.
Professor Blanco-Silva explained this to me during class last week. All you have to do is plug in the first number, like x=-1, for the first two equations. If they are the same, then it is continuous, if not then it is discontinuous at that point. Then you move on to the next number, which is x=0. You plug that in for the second and third equation and follow the same steps. This is all that you do to determine if it is continuous or discontinuous. Hope this helps.
Thank you! That helps alot
I believe that question 14 can be solved using substitution, so where’s the best place to sub the ‘u’ into the problem?
for question 14.. the best idea would be to separate into two integrals… integral 2/1+x^2 dx and integral 3x/1+x^2 dx…then in the first integral pull out the 2 so its 2 integral 1/1+x^2 and the other is still the same. the antiderivative of 1/1+x^2 is tan^-1(x) so then it would be 2tan^-1(x) cause you pulled out the 2. for the 2nd part of the integral the antiderivative of 3x/1+x^2 dx is to choose the u=1+x^2 then the du=2x dx..leaving it to be 3/2u du…3/2 ln abs(u)…the whole answer would be 2tan^-1(x)+3/2 ln abs(1+x^2)
hope that helps!!! it was hard to type out but hopefully you can understand it from this..
@travis,
Yes, use u-substitution. Try the problem with the denominator as ‘u’;
u = (1 + x^2) * dx
I do also believe that the best way to solve number 14 is by breaking it into 2 integrals with the same denominator. Once broken up, you can move the 2 and the 3 out in front of their respective integrands. By doing this, you are left with 1/1+x^2 for your first integral and x/1+x^2 on your second. The first one gives you an arctan of u and the second gives you the ln abs(U). Just integrate and plug in your numbers and you should be good!
Anyone know how to compute question 12? the first step is to see if you can plug in the limit as x goes to 0. If you plug in 0, you get 0/0 which does not work. From there I know that we must do something else to the limit. Do we take the Derivative of both the top and bottom and then plug the 0 back in for x (L’ Hopitals rule)? if so, what did you all get for the answer? Im getting that the answer is 1/6? Also, when do you know when to stop taking the derivative of one of these guys?Isnt there certain final outcomes we are looking for and certain ones we know that we have to keep going ( like after plugging in you get 0/0 you know that you have to keep taking the derivative).
any help would be great and dont forget to post any questions onto either exam so we can all help eachother out.
when working out a problem such as this, you can use l’hopitals rule as many times as necessary until you are able to plug the limit into the equation
for this problem, i had to use l’Hopitals rule three times.. you use it three times because after third time, you are able to plug 0 into lim x->0 of cos(x)..
i would look through your work and check possible sign errors because your answer of 1/6 is very close
yes, you can use L’hopitals rule multiple times. Remember that if you take the limit and your answer is 0/0 or infinity/infinity (-infinity/-infinity) you can use L’hopitals. For this particular problem you will need to do L’hopitals multiple times. (hint: derivative of -sinx is -cosx) hope this helps!
Yes L’Hospitals Rule is the way to go. dydx = (cosx – 1)/ x^3. Take the limit and still get an indefinite integral, so L’Hospital Rule again. d^2y/dx^2 = -sinx/3*x^2. Again the limit is indefinite integral, so repeat L’Hospitals Rule. d^3y/dx^3 = -cosx / 6. Take the limit as x approaches zero and the answer is -cos(0) / 6 == -1/6
For question # 9, is the answer 100/25pi? I looked it up and since radius is a constant multiplier the height is the only thing changing.
you are correct to get the answer of 100/25pi for how quickly the surface is rising, but don’t forget the units of m/s.
the radius is constant and that was a key component is solving this related rates problem. Because this is a related rates problem, remember that two things have to be changing so height is not the only thing changing. the volume is also increasing with the change in height.
David-
For question 9, I got the same value (100/25pi). I looked over the problem and the point of confusion I had and maybe your having aswell is when it states “surface rising”. I believe that the question is asking for the rate at which “h” or the height is changing, NOT the surface area. that being said, the only equation that I came up with to utilize to come up with a solution is the volume equation (v=pi*r^2*h). Once you plug in the radius (since its not changing like you said), take the derivative and plug in what we know (plug in the value of dv/dt which is 100). so we have all parts to the equation except for dh/dt which is what we must solve for to find the “how fast is the surface rising”. Your answer matches what I got for the answer.
Hope this helps confirm your answer.
For number 9, do you not use h=7m for anything?
Arran you do not since you are taking the derivative of the volume formula and it does not require h since it becomes dh/dt
Oh ya, dont forget the units!! 100/25pi m/s.
How do you compute the limit for number 2? i tried simplying then l’hospitals but im not sure my answer is right?
I believe you had the correct approach
I simplified with the common denominator of x^2(x^3+x^2)
i then used l’hopitals rule three times resulting in the function:
6/60x^2+24x
Remember the three rules Sameed gave us in recitation for limits of a polynomial over a polynomial as x->0
1. If the degree of the numerator is greater than the degree of the denominator then the limit =0
2. if the degree of the denominator is greater than the degree of the numerator then the limit=INFINITY
3. if the degree of the numerator and the denominator is he same then the limit is the ratio of the leading coeficients
Using rule 1 for the function above the answer would be +INFINITY
You don’t need to use l’hospitals for it, what you should try is using least common denom to subtract the two fractions, simplify, then since it’s asking for the lim x->0 from the left, plug in negative numbers very close to 0 i.e. -1/2, -1/4, etc. And use those values(Which are y values on the graph) to see if x is going up, down, or towards the point and up is infinity, down is -infinty, and if it can work at the point then it’s the point(Hint: It doesn’t).
A simple way is to make a chart of values that are less than zero. f(0.1), f(0.01), f(0.001). After you get these values, you can determine where the limit is going.
Sorry, f(-0.1), f(-0.01), f(-0.001). Use negative values because the limit is approaching zero from the left.
Since the problem is asking for the limit going to zero from the left, I chose -0.1 as my x value to plug into the equation. Once the value was plugged in, the equation appeared as so:
1/((-0.1)^2) – 1/((-0.1)^3+(-0.1)^2)
When simplified, this appears as:
1/0.01 – 1/0.009
What this means is that there is going to be a smaller infinity minus a larger infinity, this will result in the answer being Negative Infinity.
For 10 on the Sign line after you get your values do you plug them into the derivative or the original problem?
i am confused about the sign line but…once you find your derivative and set it equal to zero you get one candidate and the other two candidates come from the interval
are you asking if you plug these three candidates individually into the original function to find max and min? if so the answer is yes
Maximum and Minimum values are from the first derivative. Extrema that is used for inflection points are from the second derivative. For the sign line, use the first derivative. Remember to have boundaries for the endpoints!
For number 9, I know that this is a related rates problem. We are looking for the volume, which is V=Pi*r^2*L. The radius is given, and so i just plugged that in and I’m getting no where. Someone HELP!
For number 9 they also give dv/dt which is 100m3/s. So then i’m pretty sure you take the derivative Pi*r^2L of the function and plug in your two known values and solve for dL/dt.