Worksheet Squaring This Equation to Get Rid of the Square Root, it Becomes

Worksheet 0 Solutions 1. Squaring this equation to get rid of the square root, it becomes 2x2 − 11x − 26 = x2 Moving the x2 term over and factoring, we then get (x − 13)(x + 2) = 0 giving our apparent solutions to be x = 13 as well as x = −2. However, we realize that x = −2 is an extraneous solution as it does not work in the original equation so only the former is a solution. This occurs because when we squared the equation two quantities which were not equal suddenly seemed like they were. This happened in part because x2 is not what we call a one-to-one function. For instance cubing an equation would not create such an issue. 2. We note that since f is an odd root and g is a polynomial, that f and g are individually defined at all real numbers. So as long as the x value we pick exists in h, that value will be in the domain of the given composition. Since h is an even root which is actually in the denominator (as the power is negative), h is defined only at positive numbers, as negative numbers would create a negative under an even root while 0 would create division by 0. Thus the domain for the composition is (0, ∞). Note that if one tries to compute the simplified function f ◦ g ◦ h, that can be a bit dangerous as that would give us f ◦ g ◦ h(x) = f (g(h(x)) = f g x −1 6


−2 = f x−2 = x 3 which would seem to have domain all nonzero real numbers. The problem is when we perform that simplification we change the domain within that calculation which can be an issue, it’s the sort of thing that can create extraneous solutions like in Problem 1. For instance something simple like eln x = −5 can clearly not have any solution since e to any real number is positive. But if one uses the natural simplification that eln x = x, it will suddenly seem like this equation does have a solution. So have to be careful! √ m 3 where 3. Suppose to the contrary that 6 + 5 7 was a rational number n m and n are integers, n ̸= 0 (that is the exact definition of a rational number). Then we know that √ m 3 5 7= −6 n which means √ 3 7= m n − 6 m − 6n = 5 5n m − 6n Now, is a rational number as the numerator and denominator 5n are integers with the denominator still √ nonzero since n ̸= 0. Thus we can √ 3 3 conclude from this, since it is equal to 7, that 7√is a rational number. 3 But this is a contradiction since we were given that 7 is irrational. Thus our original assumption that the number in question was rational was incorrect so it therefore must be irrational, as desired. Note that an argument of this nature is really probably your only good option, you cannot do this by testing rational numbers one by one and showing that none of them work. The big issue is that if we say asked if there is an integer n for which n2 = 50, you fairly quickly would be able to say the answer is no by testing 7 and 8, realizing neither of them works and then via ordering you could establish that no integer would therefore work. However, that type of argument can’t work here because between any two distinct rational numbers there are actually infinitely many other rational numbers. So even though both the integers and rational numbers are infinite sets, they are very different structurally! Also note that the reasoning given makes it very clear why it is impossible for the given number to be rational meaning by default it has to be irrational. It’s important to be precise √ though - often times in this style of 3 question, I see people say things like “ 7 is irrational and multiplying by √ 3 5 doesn’t change that so 5 7 is irrational as well”. And while that might technically be true, it doesn’t really make things clear, that’s definitely a bit of a jump. √ After all √ if one were to say “ 2 is irrational and squaring won’t change that so ( 2)2 is also irrational”, they would clearly not be correct! So again, it’s important to make things clear, don’t make a statement which seems to be true (and perhaps actually is true) without properly justifying it. 1 4. We start with the well known graph of g(x) = and then via reflecting g x −1 over the x-axis we change this into h(x) = which gives us the graph x y x We then vertically stretch h by a factor of two to obtain the graph of −2 j(x) = x y x And finally we shift j upwards by three units to obtain the desired graph 2 for f (x) = 3 − to be x y y=3 2 3 Note that this is not the only way certainly to get the answer, for instance we should shift the order of the first two transformations. Also, if you are comfortable with this, it is certainly reasonable to make those first two transformations simultaneously to cut down the amount of work. Just need to be a bit careful when making multiple transformations at once, particularly if those transformations happen to be non-commutative x (order matters). Since that is not the case here, the first two are commutative, it is fairly safe to do that. Note that this, at least at this stage of the class, is the appropriate way to do the problem (ok there are multiple ways you can go about the transformations, but this general idea is what should be used). What is really not appropriate is to just pick a few standard x-values to obtain a few points and then “connect” those to try to create the graph in question - that is not precise! Naturally any graph of this sort has infinitely many points, and if you try to obtain the graph based upon say only three or four randomly picked points, again, you are making a lot of unsafe assumptions, that simply isn’t going to be enough to ensure your graph is reliable, particularly if the function is nontrivial. Ok, for nontrivial functions, modifying a well known graph using typical transformations may not always be possible (that’s one reason why we will later use calculus to help us graph things which are tougher), but again at this stage that is what you need to do. Not that it’s a bad idea to pick a few specific points to help yourself, but understand that alone is not enough to really create a reliable graph. 5. We wish to have x2 1 1 + 2 =0 + x + 3 3x − 9x + 1 Combining these into one fraction using a common denominator, this becomes (3x2 − 9x + 1) + (x2 + x + 3) =0 (x2 + x + 3)(3x2 − 9x + 1) which simplifies to 4(x − 1)2 =0 (x2 + x + 3)(3x2 − 9x + 1) So our only solution is clearly x = 1 as that is the only value where the numerator becomes 0 which is what we need (the denominator is not zero there so this is a safe solution). Note that one thing to be avoided in this problem is doing unnecessary calculation - that is, you should not multiply the two terms in the denominator out, that simply doesn’t accomplish anything. Since the expression is to equal 0, the denominator is irrelevant other than the fact that it should not equal 0 as well so there is no reason to perform such “simplifications” on it. More generally, whenever you are thinking about multiplying something of that sort out, or the reverse, that is factoring some expansion, try to think ahead as to how that will actually help you in the overall problem, don’t do that purely instinctively. As we can see, in this problem factoring out the numerator was very helpful towards getting the solution quickly while the denominator, whether it was factored or not, had no impact on how we proceeded. So it very much depends on the situation which again highlights why you should try to think ahead as to if what you are doing will actually help you in solving the problem or not. 6. We first work on the “stacked” fractions to write this as a single fraction to make it easier to work with. Doing that, this first simplifies to x2 · (x3 )3 · x−5 (x−2 )−7 Then we apply the same exponent rule for (xn )m to the top and bottom to further rewrite this as x2 · x9 · x−5 x14 We then use the rule xm · xn = xm+n to turn this into x6 x14 and finally we utilize xm = xm−n to make this n x x6−14 = x−8 So therefore n = −8 is our answer. In this question we had to use a variety of different exponent laws in addition to a bit of ideas of simplifying fractions to get this done. And this has to be done quite carefully - there is a lot of opportunity to make a mistake in this problem if you aren’t careful. That is without a doubt a big thing I see consistently in this class, when folks are solving a calculus problem there are often ideas involving working with exponents in addition to the calculus ideas on top. And it is imperative that you be comfortable with these exponent laws, to not make mistakes super consistently as that will really add up, especially considering how often exponents do come up in this course. 1 7. First note that the triangle in question clearly has area . Thus we wish 2 to find a vertical line, x = d say, for which the triangle that comes about 1 due to this line and the right vertex (1, 0) has area . To illustrate, we 4 have this y x=d 1 (d, ?) d 1 1 So we want the area of the yellow region to be . To that end, we compute 4 the length and width of that right triangle (in terms of d) in order to solve for d. The base of that triangle is clearly 1 − d since it is a horizontal segment from (d, 0) to (1, 0). For the height, we need to figure out the y-coordinate of the upper point (currently denoted as ?), and that will logically be the x height of the triangle. To that end, the equation of the hypotenuse of the original triangle, since it passes through (1, 0) and (0, 1), has slope m= 1−0 = −1 0−1 and therefore has equation y =1−x Hence it intersects the line x = d at the point (d, 1 − d) meaning 1 − d is 1 the height of our triangle. Thus in order that this triangle have area we 4 must have 1 1 (1 − d)(1 − d) = 2 4 which simplifies to (1 − d)2 = 1 2 1 and thus 1 − d = √ (we must pick the positive root as otherwise we 2 would have d > 1 which does not make sense geometrically, or equivalently we know that 1 − d must be positive since it represents a length). And therefore we finally get 1 d=1− √ 2 so x equal to that quantity is our desired vertical line. 8. We note that the expression in question becomes 0 at x = 1 and x = 3 while being does not exist (dne) at x = 2 and x = 4. Thus those four points are the points of relevance. Realizing this, we break the real line into five pieces with these four points being the endpoints and test the expression in question in each of those individual intervals. We find that for x < 1 since three of the four terms are negative, the overall expression is negative. Continuing like this and testing the other intervals (checking as to whether there are an even or odd number of negative signs), overall we find the sign chart to be − − + − − − − + + 1 + − − + + 2 + − + + + 3 − + + + + 4 (x − 4)3 x−3 (x − 2)2 x−1 + Therefore our answer is (taking special care as to which endpoints work and which don’t, just via plugging them in) [1, 2) ∪ (2, 3] ∪ (4, ∞) It is important to realize in this question that we cannot simply cross multiply by the denominator as we might be tempted to do in an equation (like we might have done in Problem 5). This is because the denominator is not irrelevant here as it mostly is in an equation (avoiding it being 0 is the only thing we have to worry about in an equation). But in an inequality, when comparing to 0, the sign of the denominator is just as important as the sign of the numerator so we do not treat it any differently outside of the fact that the points where the denominator is 0 are not defined while the points where the numerator is 0 make the expression 0, that is, the expression is still generally defined at the latter type of points. You have the option in this situation to, as I have done here, test/create a sign chart for each factor individually and then assess the overall based on that. Another option is to just test the entire quantity at once. The latter method is a bit faster but probably easier to make a mistake in as well. So with fewer factors I would almost certainly use the latter method while with more (such as this case), a bit less clear. You can do whichever you feel more comfortable with though - just of course make sure it’s correct! This is a bit ahead of where we are, but to make a point: a rather important concept that we deal with in regards to functions is wanting to know where a function is what we call increasing vs decreasing. As it turns out, the main way to do that is using what we call the function’s derivative (calculus term) and seeing where it is positive vs negative. That is, to figure out where a function is increasing, you mostly must figure out where its derivative is positive, i.e. solve the derivative > 0. So in short, it boils down to solving an inequality to figure that out. And that’s one reason why I think it’s important for you to practice that now. The simple fact is, if you want to figure out where a function is increasing vs decreasing (which will 100% be a concept that will come up later, reasonably often perhaps), if you can’t solve inequalities, you won’t be able to figure that out! So that’s a very relevant thing to know how to do going forward.