Calculus AB / Mr. Hansen

Name: __________________________________

Explanations for Cumulative Review Quiz, 1/9/2004

1. _E_	Since limit is of form 0/0 (indeterminate), apply L’Hôpital’s Rule.
2. _B_	This is a precal review problem. Since f (2) = 8 + 4 = 12, plug in to get g(8) = Ö(12 – 11) = Ö11.
3. _C_	This is similar to #2, but students found it much harder. Since f (1) = ln 1 = 0, plug in to get g(0) = e^3 · 0 = e^0 = 1.
4. _E_	Yet another precal review problem, this one requires checking whether (A) y and –y are interchangeable, (B) x and –x are interchangeable, (C) (x, y) is on the graph iff (–x, –y) is on the graph, or (D) (x, y) is on the graph iff (y, x) is on the graph, i.e., x and y are interchangeable. A quick glance reveals that none of these conclusions are satisfied. Or, if you are good with algebra, you can complete the square twice and perform some manipulations to get the equation    (x – 3)^2/25 + (y + 1)^2/16 = 1. In both Algebra II and Precalculus, you learned that this is the equation of an ellipse with center at (3, –1), horizontal semi-major axis of 5, and vertical semi-minor axis of 4. A quick sketch reveals that none of the symmetries described in (A) through (D) are satisfied.
5. _E_	Apply FTC: ò2^7 x^3 dx = x^4/4 |2^7 = 7^4/4 – 2^4/4 = 49^2/4 – 16/4 >> 40^2/4 – 4 = 1600/4 – 4 = 400 – 4 = 396. Clearly, (E) is the only reasonable choice. On a “no calculator” portion of the test, the numbers will seldom get this large, but if they do, you should be prepared to use some estimation shortcuts as shown above to save time. Here is a way of getting the exact answer relatively quickly without a calculator: ò2^7 x^3 dx = x^4/4 |2^7 = 7^4/4 – 2^4/4 = 49^2/4 – 16/4                  = ¼(49^2 – 4^2) = ¼(49 + 4)(49 – 4) = ¼(53)(45)                  = ¼(53)(40) + ¼(53)(5) = 530 + ¼(265) = 530 + ½(132.5)                  = 530 + 66.25 = 596.25 There is also a faster way, if you are clever enough to do get inside the head of the question writer. You should already know that 7^3 = 343. (The number 343 also has significance as the number of FDNY heroes who died on 9/11/01.) Since 7^3 = 343, you know that choices (C) and (D) are meant to fool students who thought they could compute the integral as 7^3, or as 7^3 – 2^3. Choices (A) and (B) are much too small, since a trapezoid rule estimate of ò6^7 x^3 dx all by itself would be ½(6^3 + 7^3) » ½(200 + 350) » 275. By the time the curve y = x^3 has reached 6 or 7 on the x-axis, the curve is practically going straight upward, and this estimate of 275 is surprisingly accurate. (Use your calculator if you doubt this.) Thus (E) is the only reasonable choice.
6. _B_	By implicit differentiation (review if necessary!), we have    6x – [2xy^3 + x^2(3y^2y¢)] + 4y¢ = 0    –3x^2y^2y¢ + 4y¢ = –6x + 2xy^3 y¢(–3x^2y^2 + 4) = –6x + 2xy^3 Divide both sides of this last equation by (–3x^2y^2 + 4) to get the answer. Note that the two tricky parts are (1) getting everything involving y¢ on the LHS of the equation and (2) factoring out y¢. If you can remember these steps, and if you can accurately take all the derivatives without forgetting the chain rule, then you can do any problem of this type.
7. _E_	We require that the limit from the left, namely 3(–1)^2 + 2, must equal the limit from the right, namely –c(–1) + 5. In other words, we require    3 + 2 = c + 5 By inspection, c = 0 is the only value of c that works.
8. _D_	Let u denote the radicand, i.e., (8 + x^2)/(x^2 + x). Since the cube-root function is continuous, the limit of the cube root of u must equal the cube root of the limit of u. We can find the limit of u either by inspection or by L’Hôpital’s Rule. Method 1 (inspection): Since numerator and denominator are both of degree 2 and have the same leading coefficient, lim u = 1. Method 2 (L’Hôpital’s Rule): Since u is of the indeterminate form ¥/¥, lim u = lim [2x/(2x + 1)] = lim [2/(2 + 0)] = 1. Note that we actually applied L’Hôpital’s Rule twice here. Either way, we know that lim u = 1. Taking the cube root of this limit gives a final answer of 1.
9. _D_	We discussed this a number of times in class. The “V”-shaped graph of f (x) = |x| has a cusp at the origin but is linear (hence differentiable) everywhere else. WARNING: The TI-83 and TI-83 Plus calculators both return the wrong answer for f ¢(0), because they use the symmetric difference method for estimating the numeric derivative. They should give an error message, but instead they will happily tell you that nDeriv(abs(X),X,0) equals 0. The TI-89 does not have this bug.
10. _B_	By the hint, (1, 4) is on the graph of f. Therefore, (4, 1) is on the graph of f ^–1 (also known as g). Since x and y trade roles in f and g, asking about the instantaneous slope of g at (4, 1) is astonishly similar to asking about the instantaneous slope of f at (1, 4). Since f ¢(x) = 10x^4 – 3x^2 + 2x, the instantaneous slope of f at (1, 4) is found by computing f ¢(1) = 10(1)^4 – 3(1)^2 + 2(1) = 10 – 3 + 2 = 9 Is that the final answer? No, because in f the function is climbing steeply, but at the mirror image point (4, 1) on the g curve, the roles of x and y are interchanged. Thus, the rise/run for f would be the run/rise for g. To find the rise/run for g, we must take the reciprocal of 9. The final answer is 1/9. Your textbook gives a confusing formula on p. 153. (Maybe it makes sense to you. If so, feel free to use it. However, I have always found it easier and safer to think through the steps as described above.) This is what the formula says:    If g = f ^–1, then g¢(x) = 1/f ¢(y). (I prefer this version to the extremely confusing version given in your textbook.) If we apply the formula to our problem, we see that the question concerns the point (4, 1), i.e., where x = 4 and y = 1. Therefore, g ¢(4) = 1/f ¢(1) = 1/9. The only other work you would need to perform is the calculation of f ¢(1) = 9 as shown above.
11. _E_	Since f ¢(x) = 6x^2 – 18x + 12 = 6(x^2 – 3x + 2) = 6(x – 1)(x – 2), the sign of f ¢ is determined by the factors (x – 1) and (x – 2). A number-line sketch (remember those from Algebra II and Precalculus?) reveals that the factors are usually either both positive or both negative. Only for x Î (1, 2) do we have one of each sign, making the product negative.
12. _E_	Since f (x) = (2 – x)^½, we have f ¢(x) = –½ (2 – x)^–½. For values slightly smaller than 2, we have    –½(very small positive number)^–½ = –½/(square root of very small positive number) = –½ · (huge positive number).