Math 123

Here are some solutions to Prof. Kofman's 2013 midterm 2 that we worked on yesterday.

I said this in class, but just to write it down here, the exam tomorrow will cover all the material up to and including Section 4.6.

Here are the solutions to today's quiz.

Just a warning to everyone that today is the drop deadline.

Our second midterm will be held on Wednesday, November 19th. The basic format will be the same as the first midterm. You're allowed a notesheet with the same rules as before. Probably the exam will cover material through Section 4.6, but I'll confirm that next week.

The best way to study is do exam-like problems under exam-like conditions. There are some good sample problems in the midterms in the Exam Archive that I posted before (since that midterm was held a bit later). I've posted some sample final exam problems there as well, now, some of which are good for this midterm too. Most usefully, I've added a link to Prof. Kofman's old website for his MTH 123 class, which contains many old exams. The more you practice with exam problems, the better you'll do.

We have another quiz coming up on Monday, November 10th. For the quiz, as usual it will be on the homework due today. Some of the homework is a bit trickier, so let me focus things by saying that there will be only one question, and it will be drawn from Lesson 24 (Modeling with Quadratic Functions).

Here are the solutions to today's quiz.

Office hours this morning are delayed till approximately 11:05am this morning because I have to do a course observation in our department. This is the last one of the semester so after this office hours should be normal again. Alaso sorry for the repeated announcements on Brightspace and here, I just wanted to make sure everyone got the message.

I wanted to give some comments on the exam problems, besides their solutions (version A and version B). On #1, performance was generally strong on (a) and decent on (b). Most people on (b) realized that they needed to solve two different equations (\(3(x-4)=8\) and \(3(-(x-4))=8\)) but lots of people got it wrong what the two equations were. One mistake was to mess up the parentheses and solve \( 3x-4=8 \) instead of \(3(x-4)=8 \). Another was to solve \( 3(x-4)=8 \) and \( 3(x+4)=8 \). The best way to think about it is that \( \lvert x-4\rvert \) is either going to be equal to \( (x-4) \) or to \(-(x-4)\), so you need to replace \( \lvert x-4 \rvert \) by these two expressions and solve both resulting equations. I was also generally happy with the way people wrote their solutions, as a list of equations, one after another.

While #1 was an exercise of solving equations, problems #2 and #3 were exercises in transforming equations. This was from earlier in the class, and probably this is why people were less likely to write their solutions in the right way, as a chain of equalities. I say this all the time, but let me say it again: When you're solving equations, you write down a list of different equations, each derived from the last. The different terms in all these equations are not all equal to each other, so you don't get a single chain of equations. When you're transforming expressions into different forms, each transformed expression is equal to the last one. So, you get a single chain of equalities. You can write this chain of equalities all on one line, if it fits, or you can break the line at some point and start the next line with an equal sign. (Typically, people make the equal signs line up, because it looks nice and is easy to read.) One thing I saw people do a few times was to write arrows from one expression to the next. I appreciate that you were trying to indicate that the expression had been transformed into another one, but an arrow is a mathematical symbol already (it's used in a few different ways, none of which comes up in this class) and it does not mean this. The right symbol is an equal sign. (If you wanted to indicate that there's something different about the situation than when there is an equation to be solved, the right way to do it would be to preface the equation with, "For all x". That's what makes the difference: when you're solving an equation, the equation is only true for specific values of x and you're trying to figure out which. When you're transforming expressions, the equation holds for all choices of x.)

In 2(b) and 2(c), one step of the problem was to square an expression, computing \( (x^3+x^{-2})^2 \) in (b) and \( (\sqrt{x+2}+4)^2 \) in (c). It was very common for people to do this by just squaring each term separately, but this is wrong because it is not true that \( (a+b)^2=a^2 + b^2 \). I have to be harsh in my grading here, because nobody should pass an algebra class and make this mistake. The correct version is \( (a+b)^2=a^2+2ab+b^2\), which in 2(b) gives you \[ (x^3+x^{-2})^2 = (x^3)^2 + 2x^3x^{-2}+(x^{-2})^2 = x^6+2x + x^{-4} \] and in 2(c) gives you \[ \begin{aligned} \bigl(\sqrt{x+2} + 3\bigr)^2 &= (\sqrt{x+2})^2 + 2\sqrt{x+2}(3) + 3^2\\ &=x+2 + 6\sqrt{x+2} + 9 = x+11+6\sqrt{x+2}. \end{aligned} \]

In 2(b), using the work above, we have \[ (x^3+x^{-2})^2 + x = x^6+2x + x^{-4} + x = x^6+3x+x^{-4}.\] The problem statement says to get rid of all negative exponents, which means writing \( x^{-4} \) as \( \frac{1}{x^4} \), giving a final answer of \[ x^6+3x+\frac{1}{x^4}. \] There was a very common mistake in this step, which was a bit mystifying to me, of instead putting \[ \frac{x^6 + 3x}{x^4}. \] In this answer, instead of adding \(x^{-4}\) to \(x^6+3x\), we've multiplied \(x^{-4}\) by \(x^6+3x\).

Problem #4 was basically the same problem as on the quiz. It reduces to the problem of finding height and width that multiply to 2000 (or 1800 in version B of the exam) and add to 100. The majority of people got to this point. Unlike on the quiz, I took care to make it so that you couldn't get the right answer just by going through the factors of 2000. Instead, you need to use the two equations \(xy=2000\) and \(2x+2y=200 \) to get a quadratic equation, which you have to solve using the quadratic formula.

People did well on problems #5 and #6. The only common mistake on #6 was messsing up the ordering with the negative slope, which is the smallest. (Negative numbers are smaller than positive ones.) I guess you understand lines well, which is good.

For #7, you need to know what the domain of a function is: the values that you can plug into it. Here, the only reason you wouldn't be able to evaluate the function is if the denominator is zero, which happens when \( x^2+3x =0 \) in version A of the problem. You have to solve this equation, which you can do by factoring to obtain \( x(x+3)=0 \), giving the solutions \( x =0\) and \(x=-3\). Then the domain consists of all real numbers except these two. You can just write that sentence. Or you could write \( (-\infty,-3)\cup(-3,0)\cup(0,\infty) \), which is perfectly valid if perhaps a bit less clear. (If you really want to write the answer purely in mathematical symbols, the best way would be to write \( \mathbb{R} \setminus \{-3,0\} \). The symbol \( \mathbb{R} \) means the real numbers, and the symbol \( \setminus\) is an operator along the lines of \(\cup\) or \(\cap\), but it means to remove the second set from the first. So \( \mathbb{R} \setminus \{-3,0\} \) means the set of real numbers with -3 and 0 removed.)

Problem #8 was probably the one that people did the worst on. This was illuminating to me, and I'll talk more in class about graphing and about piecewise functions. Problem #9 was not exactly easy, but it was the same basic problem we practiced a few times before the exam.