+ i sin. + i sin. = 2 cos

Transcription

1 Math 11 Lesieutre); Exam review I; December 4, a) Find all complex numbers z for which z = 8. Write your answers in rectangular non-polar) form. We are going to use de Moivre s theorem. For 1, r is 1 and θ = π. We want n =. The solutions are then going to be: z = ) )) π π 8 cos = cos + πk ) π + πk + πk )) π + πk The range here is k = 0, 1,. So are points are: z = cos + 0 π ) π + 0 π )) ) π )) = cos = 1 + i, = cos + 1 π ) π + 1 π )) = cos π) π)) =, = cos + π ) π + π )) ) )) 5π 5π = cos = 1 i. b) What are all the solutions to z 4 = 16i? You can leave your answers in polar form.) It s the same game. This time, a = 16i. The point has radius 16 and θ = π. We want to use n = 4. That s everything we need to know to use the theorem... z = 4 / 16 cos 4 + πk ) π/ πk )) 4 = cos 8 + πk ) π 8 + πk )) Since n = 4, we are allowed to use k = 0, 1,,. The options are: z = cos 8 + 0π ) π 8 + 0π )) = cos 8 + 1π ) π 8 + 1π )) = cos 8 + π ) π 8 + π )) = cos 8 + π ) π 8 + π )) Let s just leave it in polar like that. 1

2 . Suppose you have a polynomial with zeroes at 1, 1, and, with multiplicities,, and 1, and that f0) = 7. What could be the formula for your polynomial? We know it has to be fx) = ax + 1) x 1) x ) We need to find the leading coefficient a. To figure it out, plug in 0: So 4a = 7, and a = 7 4. f0) = a+1) 1) ) = 4a fx) = 7 4 x + 1) x 1) x ). Solve each of the following equations for x. a) x+1 = That means x+1 = = 5 x + 1 = 5 x = x +1 = 9 x +1 = x + 1 = x = 1 x = 1 or 1 b) ln x = lnx + 4) + lnx + 1) lnx) = lnx + 4) + lnx + 1) lnx) = ln x + 4)x + 1)) x = x + 4)x + 1) = x + 5x + 4 x + 4x + 4 = 0 x + )x + ) = 0 x =, and that s the only solution.

3 c) e x = e x Let y = e x. Then this is just y = y, so y y = 0 and yy 1) = 0. Thus y = 0 and y = 1. The former means e x = 0, which has no solutions. The latter means e x = 0 so that x = 0. That s going to be the only solution for this one. d) cos x + sin x 1 = 0. This is a similar idea to c), but with trig functions. First turn cos into 1 sin : cos x + sin x 1 = 0 1) 1 sin x + sin x 1 = 0 ) sin x sin x = 0 ) Le u = sin x. Then u u = 0, so u = 0 or u = 1. The former gives x = 0 + πk or π + πk. The latter gives π + πk. 4. Sketch graphs of the following functions. a) fx) = x 1) x + ) what is the end behavior?) The key things to notice are that there is a root of multiplicity at x = 1 so it bounces off the axis), and a root of multiplicity 1 at. The end behavior is that y when x and y when x. Here it is: b) gx) = 1 x what is the inverse function?) Think of this as x+1) We start with x, which is one of the basic graphs that you should know. x reflects that about the y-axis. Turning the x into an x + 1 then shifts it left by 1. The is a vertical stretch. The result is the the following.

4 You could also think of this as being 6 x, so you don t have to deal with a horizontal shift. To get the inverse function we switch x and y to get: x = 1 y x = 1 y 1 y = log x/) y = 1 log x/) c) ix) = log x 1) + 4 This is another secret transformation problem. You hopefully know basically what log x) looks like see the worksheet we plotted it). log x) : start here log x 1) : shift right 1 log x 1) + 4 : shift up 4 4

5 there s a vertical asymptote at x = 1 that doesn t show up very well in this picture) 5. A fly colony starts with 100 flies, and three days later the number has risen to 00. Assuming that the colony exhibits unconstrained exponential growth, find a formula for the number of ants after t days. When will there be 1000 flies? Unconstrained exponential growth means that Nt) = N 0 e kt, where N 0 and k are some constants. Here N 0 is the initial number of ants, which is 100. To find k, we need to plug in for t, and use the fact that N) = = 100e k = e k k = ln k = 1 ln ). So our formula is 100e 1 ln)t = 100e ln ) 1 t = 100 t/. When will there be 1000 flies? It s when 100 t/ = 1000, so t/ = 10. Then t/ = log 10) and t = log 10). 6. Consider the rational function fx) = x 1)x )x ) 5x x 1) 5

6 a) What are the vertical asymptotes of this function? Lucky for us, it s already factored; otherwise, that would be the first thing that you want to do. We notice: there s an x 1 both top and bottom. So for everything except finding holes, we use the simplified formula x )x ) 5x Just don t forget about the hole at x = 1 when you go to draw the graph. b) Does the function have a horizontal asymptote? If yes, what is it? Numerator and denominator have the same degree, so there s a horizontal asymptote, at y = 1 5. c) Sketch a graph of the function. There s a vertical asymptote at 0, and it s 0 at x = and x =. Next we want to go through all this rigamarole of where it s positive and where it s negative to help us draw a graph. Now we want to figure out where it s positive and where it isn t. We split it into intervals, with the breakpoints at x = 0, x =, x =. interval test value sign, 0) 1 + 0, ) +, ).5, ) 4 + Putting that together, we get the following plot: I didn t show the hole on this one, but it s there. Also notice the horizontal asymptote at y = 1/5. It s a little hard to see the x-intercepts at and, but they re there.) 6

7 7. You start with $00 in a bank account paying 1% interest. How much money would you have after twenty years if interest is compounded monthly? daily? continuously? Remember the equation: A = P 1 + r n) nt. Plugging in for monthly, n = 1 and we get. A = ) 1 0 This is about $44.6. For daily, Plugging in, A = P 1 + r n) nt. A = ) 65 0 $ Continuously is a different formula: P e rt = 00e $ Sketch a plot: y = cosx π) + 1. You want to rewrite this as cos x π )). The amplitude A is. The phase shift is π to the right compared to regular cosine). The period is π. The midline is 1. I would graph this in two stages: first do cosx) without either the vertical or the horizontal shift), and then shift right by π and up by Expand the following expression as much as possible: Looks like this is going to be log log x + 1) x x 1)x ) 7 x + 1) x x 1 x ) 7 = logx + 1) logx) log x 1 logx ) 7 = logx + 1) + logx) 1 logx 1) 7 logx ). 7