Exam Solutions

The mean was about 90/135. The high score was 129 and the low score was 39. In general the you guys did alright on the exam. However, the responses for Problem 1 were rather dissapointing. After I specifically said to read the significant section in the book, gave a very similar problem on a quiz, when over the similar problem in class, and posted solutions on the course webpage, hardly anyone got this problem correct. For the most part, most people did well on problem three. I just want to point out that Problem 3e asks you to fined the distance from the midpoint of a line segment to one of its end points and Problem 3f asks you to find the distance between the two end points of the line segment. While it is totally fine to use the distance formula in both cases, I was hoping at least one of you would realize you could save yourself some work and get the answer to Problem 3f by multiplying the solution to Problem 3e by 2.

The mean was about 91/125. The high score was 123.5 and the low score was 55.5/125. In general the you guys did not do as well as I would have hoped but overall you did alright on the exam.

There were a few common mistakes in problem 3bi. Some people wrote f\circ g(x)=(\sqrt(x-1)^2)=x-1, which is right, but then they proceeded to set x-1=0 and claimed x=1. This is very very bad. By writng x-1=0, you are saying that f\circ g(x) is the zero function and this just is not true. You also need to make sure that you write true statements. Some people wrote f(x)=x-1 where it is f\circ g which is equal to x-1 and f(x)- x^2. Finally, people still had trouble with finding the domain of a function that looks like y=\sqrt(F(x)). The domain is all real numbers x which satisfy F(x)\ge 0.

The big mistake in problem 3bii was saying that the domain of g(x)=\sqrt(x-1) was all real number except x=1. This just is not true. For one, x-1, is in the domain and two, there are infinitely many values for x such that x-1<0, i.e such that \sqrt(x-1) is not a real number. In problem 3ci, there was a lot of bad algebra. Make sure you know how to solve algebraic equations for one variable in terms of another.

I was highly disappointed with the answers for problem 4. Not only was this a homework problem but it was a homework problem, I mentioned on the homework page and discussed in class. Even with the hint, a lot of people were unable to realize that the domain of f(x) is all real values x such that x+1/(x-4) \ge 0.

In problem 5a, I was happy to see that a lot of people were able to complete the square. In general you guys did pretty well on this one.

The only part of problem 5 a lot of people had trouble with was part (e). A lot of people said h looked like (x+2)^3 near the zero -2 and completely ignored the other factor x(x-1)^2 of h(x). Worse than that, a few people said h(x) looked like x^3 near the zero x=-2. This was worth about 1 point since it tells me you know that the behavior of h near the zero -2 depends on the multiplicity but that you don't really understand what is happening to the function h, near its zero. A lessor mistake involved misuing equality signs. A lot of people wrote that near x=-2, h(-2)=-18(x+2)^3. This statement is not true; h(-2)=0 and h is the function h(x)=x(x-1)^2(x+2)^3 . You really need to write the quantifier "near x=-2", proceeded by "h~ . . ."

In problem 7, even after we talked about it in class, a lot of you still thought that vertical asymptotes were zeros and zeros were vertical asymptotes. VERTICAL ASYMPTOTES ARE NOT ZEROS. In a rational equation R(x)=p(x)/q(x), which is in lowest terms, the zeros of R(x) are precisely (only) the zeros of p(x) NOT the zeros of q(x). The zeros of q(x) are not even in the domain of R(x). In part f of problem 7 a few people mixed up the definiation of a function with the definition of one-to-one. Make sure you know the difference.

The the mean score was 68/100 with the lowest score 27/100 and with the highest score 96.5/100. Overall I was pretty disapointed most solutions to most problems; so much so that I am tempted NOT to scale the scores for this exam. Make sure you do NOT repeat the mistatkes made on this exam on the final. I will be grading the Final carefully.

On Problem 1, hardly anyone got all the T/F statements correct and almost everyone got part a wrong. I should not have to tell you that log(a) - log(b) is NOT the same as log(a)/log(b) for all positve number a and b. Suppose a is 10 and b is 100. Then, log(a) - log(b)= 1-2=-1 and log(a)/log(b)=1/2 and clearly 1 is NOT the same as 1/2. If blank out pick some numbers you know how to deal with and check to see if what is written makes sense. It is really really important you guys nail you algebra skills. You make it much further in math classes if you don't. A lot of people missed part c as well. Part c is true. It is exactly the statement of the change of base formula with number filled in. If the number threw you off, you should go back and make sure you understand the definiation of a function. Also a handful of people missed part d. This is a very bad problem to miss. Missing it tells me that you don't understand what the function log_a(x) is. log_a(x) is whatever valued I have to raise a to to get x (i.e. it is the inverse function of the exponential a^y). If you just say that sentence every time you want to know a logrithmic output, it should be obvious that the output of log_2(1) is 0 NOT 1. Again if you know what the logarithmic funtion is part e should not have been a problem.

Half of problem 2 is checking to see that the possible solutions you get from solving a quadratic are in the domain of each of the logarithmic functions which appear in the equation. Many of you failed to do this. It really is important. In adddition, there was a lot of bad algebra on Problem 2. Make sure you undertand the properties of logarithmic functions. A lot of people made the two following (painful) mistakes by writing ln(x)+(ln(x-6)=ln(x)ln(x-6) and ln(x^2-6x)=ln(x^2)-ln(6x). These statemenst are NOT true. I have a few notational comments. First, make sure what your write is a true statement. If something is equal to zero you need to write "something = 0" not just the "somthing" part. Also when you write something like log(4x-16) you do really need the parenthesis. The statement log 4x-16 is unclear. It could mean log(4x) -16 or log(4)x-16 or log(4x-16). Make sure you write the parenthesis next time.

Problem 3 was not bad. A lot of you answered parts a, b and c for the function log_{1/2}(x) instead of log_{1/2}(x-3). Also some of you had trouble with part b. It seemed like you knew that 1/2 and the -3 had something to do with the points on the graph but didn't know how. The best way to do this (and a lot of you did do it this way) maybe to find 3 points on the graph of the funciont log_{1/2}(x) and then shifts those points 3 units to the right so they are on the graph of log_{1/2}(x-3).

I don't understand why there was so much confusion with problem 4 part a. This problem came word-for-word from your homework. It should NOT have come as a surprise. Your confusion with the statement of this problem suggets to me that you either did not understand the homework problem in the first place or that you didn't review your homework in stuyding for the exam. If the former is true, come ask questions about the homework while you are doing it. If the latter is true, take this as a learning opportunity; going over your homework is one good thing to do when studying for an exam. For the record, all you had to do for part a was draw a circle, label it as the unit circle, draw in some angle, label it \theta, find the corresponding point on the unit circle and label the x coordinate with the function cos(\theta) and the y coorditate with the function sin(\theta). This question is really just asking you to give the definitions of the sine and cosine funtions. For part b, when you label and angle you need to draw a semi circle to indicate where the angle is. Also you really should label this semicircle with your angle measurement, not the point on the unit circle (see test solutions). Part c should have been a quick problem if you know what it means for sin(\theta) to be an odd function and for it to have period \pi. A few of you knew this, which I was very happy to see. Because sin(\theta) is odd sin(-5pi/6)=-sin(5pi/6) and because sin(\theta) has period 2pi sin(5pi/6)=sin(5pi/6 +2pi)=sin(5pi/6 +4pi).

There were some good answers for problem 5a (the defintion of a radian). Writing that 1radian=pi/180 degrees does not give the definition of a radian. All it tells me is that you know a radian is a measure of angle. If you wrote this equation or that a radian measures angle you received one point, which I think is generous. You need to say something along the lines of: a radian is an angle measure which corresponds to an arclength of 1 unit along a unit cirle. For part b, some of you wrote down the equation for the area of a sector instead of the arclength. Also a few of you converted to degrees when calculating arclength. This is bad. The equation for arclength s= r\theta (radius times angle) is only true when \theta is in units of radians. The follows from the definition of a radian and it's dependence on an arclength of one unit on the unit cirlce. So whenever you calculate arclength makes sure you use degrees. Units are important.

Most of you did well on problem 6. A few of you did not realize that there are two cycle of sin(\theat) and cosi(\theta) in the interval [-2pi, 2pi] and some of you did not label quite enough points to convince me you understood the graphs. Other than that I was pretty happy with the solutions to this problem.

You guys did pretty well on problem 7. A lot of people made the same algebra mistake, that was to write that 2pi/(1/3pi)=6pi. 2pi/(1/3pi)=6pi^2. At this point, such algebra mistakes are really unexceptable. A few people said that the period of the tangent function is 2pi. It's not. The tangent function has a period of \pi. A lot of people had trouble with the amplitude and range of 1/8tan(4x). An amplitute does not really make sense here since the range of the tangent function is all real numbers but if you wrote 1/8 I gave you full credit. Also because the range of the tangent function is all reals, the range of the function 1/8tan(4x) is still all real numbers; multiplying an arbitarily large number or an arbitraily negarive number by 1/8 is still arbitrarily large or arbitrarily negartive. A lot people made the mistake of writing that the range of 1/8tan(4x) was [-1/8, 1/8].

Most people did well on problem 8. A few people did not get the signs of cos(\theta) and sin(\theat) correct. The sign is an important part of your solution in this problem. Some people forgot to scale, saying that since tan(\theta)=4/5 and tan(\theta)=sin(\theta)/cos(\theta), cos(\theta)=-5 and sin(\theta)=-4. There's a pretty big problem, the coordinates (cos(\theta),sin(\theta)) are always on the unit circle and the point (-5,-4) is not. The remedy this all you have to do is scale by the radis of the circle that the point (-5,-4) lies on.

Problem 9 is one of those problems that you either get or you don't. I was happy to see that most of you got it.

I was pleasantly surprise by what a lot of you wrote for the bonus problem. Although only three people got credit, from what most people wrote I could see that you were thing about what the domain means for a composition of funtions. If you want to know the solutions, see the test solutions.