Calculus with Analytic Geometry I Exam 8 Take Home Part.

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1 Calculus with Analytic Geometry I Exam 8 Take Home Part. INSTRUCTIONS: SHOW ALL WORK. Write clearly, using full sentences. Use equal signs appropriately; don t use them between quantities that are not equal. But use equal signs when needed; an arrow is not an equal sign. This being work to be done at home, points will be lost for the following common nono s : Lack of a staple=2 points. Torn pages=3 points. Excessive crossed out material = 4 points. Crumpled paper=4 points. Deemed unacceptable as work done at home = 100 points. 1. (Textbook, Section 4.1 # 34) Find the critical points of the function g(t) = 3t 4. so that We have g(t) = g (t) = { 4 3t, t < 4/3 3t 4, t 4/3 { 3, t < 4/3 3, t > 4/3 We see g is never zero, but g is not differentiable for t = 4/3. The only critical point is 4/3. 2. (Textbook, Section 4.1 # 54) Find the absolute maximum and absolute minimum of f(x) = the interval [0, 3]. x x 2 x + 1 in It should be noticed that the denominator is never 0, so the function is everywhere differentiable. f (x) = x2 x + 1 x(2x 1) 1 x 2 (x 2 x + 1) 2 = (x 2 x + 1) 2 f (x) = 0 if (and only if) 1 x 2 = 0; critical points are 1, 1, but only 1 is in the interval. testing at the endpoints and at the only critical point: f(0) = 0, f(1) = 1,f(3) = 3/7. The absolute maximum is 1, the absolute minimum is 0. Penalties: For testing f at 2; since this indicates that you are still acting as if the world of numbers is reduced to integers, and/or do not understand or believe that a local extremum can only happen at a critical point: 5 points out of (Textbook, Section 4.1 # 56) Find the absolute maximum and absolute minimum of f(t) = 3 t(8 t) in the interval [0, 8]. f(t) = t 1/3 (8 t) is everywhere differentiable inside the interval (it isn t differentiable at 0 but that is an end-point); f (t) = 1 3 t 2/3 (8 t) t 1/3 = t 2/3 ( 1 3 (8 t) t ) = 1 3 t 2/3 (8 4t). This is 0 only when t = 2. Testing at the one and only critical point and at the endpoints: f(0) = 0, f(2) = 6 3 2, f(8) = 0. Similar penalties as those mentioned for the previous exercise apply. The maximum value is 6 3 2, the minimum value is (Textbook, Section 4.1 # 70) An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle θ with the plane, then the magnitude of the force is µw F = µ sin θ + cos θ, where µ is a positive constant called the coefficient of friction and where 0 θ π/2. Show that F is minimized when tan θ = µ.

2 2 Comment: This is a so-so model of what really happens when you drag an object, say a piece of furniture, along the floor. If you pull parallel to the ground (angle θ = 0), friction is strongest; you are better off pulling a little bit in an upward direction. Of course if you pull completely upward, friction has no effect, but the object will be lifted rather than dragged along the floor. All this is not terribly relevant to solving this problem µw which could have been stated simply as follows. Show that the minimum value of F (θ) = µ sin θ + cos θ in the interval [0, π/2] occurs when tan θ = µ. 5. (Textbook, Section 4.2 # 6) Let f(x) = tan x. Show that f(0) = f(π) but there is no number c in (0, π) such that f (c) = 0. Why does this not contradict Rolle s Theorem? Rolle s theorem requires that the whole interval (a, b) between the points a, b where f(a) = f(b) be in the domain of f, and that f be differentiable at all points of that interval. In this case, π/2 (0, π) is not in the domain of f, so Rolle s theorem does not apply. For the functions given in the next two exercises, (a) Find the intervals of increase and find the intervals of decrease. (b) Find the local maximum and minimum values, and state where they occur. (c) Find the intervals where the function is concave up, the intervals where it is concave down, and the inflection points. (d) Use the information from parts (a)-(c) to sketch the graph. 6. (Textbook, Section 4.3 # 36) g(x) = x 3 + x 4. This exercise was done in full in class. Here is a graph. 7. (Textbook, Section 4.3 # 38) h(x) = 5x 3 3x 5. We compute and set to zero the derivative and second derivative: h (x) = 15x 2 15x 4 = 15x 2 (1 x 2 ) = 0 roots: 1, 0, 1. h (x) = 30x 60x 3 = 30x(1 2x 2 ) = 0 roots: 1/ 2, 0, 1/ 2. We see that 1, 0, 1 are critical points, and 1/ 2, 0, 1/ 2 are possible inflection points. One thing we might do at once is see if we can classify the critical points. We see that h ( 1) = 30 >, so we have a local minimum at 1. h (0) = 0, so 0 is undecided. h (1) = 30 < 0, so we have a local maximum at 1. I like to set up a table more or less like this one, with the information found so far. The x-values appear in order. x y Classification 1 2 cr. pt., min, y > 0 1/ 2 7/(4 2) y = cr. pt., y = 0 1/ 2 7/(4 2) y = cr. pt., max, y 0 y = 0

3 3 This table tells us most everything. For example we know that the function is concave up until at least 1/ 2 and will be concave down from 1/sqrt2 onwards. We have a minimum at 1, that means that the function increases to the right of 1 and will increase at least up to 0. Past 0 the only other critical point is at 1, where the function has a maximum. That means that the function was increasing to the left of it. So h had to increase from 1 to at least 0, but to be increasing to the left of 1 it has to be increasing from at least 0. The conclusion is that the function increases from 1 all the way to 1, making 0 an inflection point. A bit of geometric intuition also shows that if 0 is critical and an inflection, the function must have been concave down to the left of 0, up to the right. Think about it. Since it was concave up at -1, concavity must have changed before 0, and the only place it can change is at 1/sqrt2. This makes 1/ 2 an inflection point. Similarly, 1/ 2 is an inflection point. It is now easy to draw a reasonably good graph. To make it even better we can find the x and y intercepts. Well, we have the y intercept already; for the x-intercepts we set y = 0; i.e., 5x 3 3x 5 = 0. The equation one gets could be very hard to solve; here it isn t. The solutions are 0, ± 5/3. So we now have our complete x y Classification 5/ x-intercept table 1 2 cr. pt., min, y > 0 1/ 2 7/(4 2) y = 0, inflection 0 0 cr. pt., y = 0, inflection 1/ 2 7/(4 2) y = 0, inflection 1 2 cr. pt., max, y 0 y = 0 5/ x-intercept The answers to points (a)-(c) are (a) Find the intervals of increase and find the intervals of decrease. The function increases in ( 1, 1). It decreases in (, 1) and in (1, ). (b) Find the local maximum and minimum values, and state where they occur. There is a local minimum at 1; the value is 2. There is a local maximum at 1, the value is 2. (c) Find the intervals where the function is concave up, the intervals where it is concave down, and the inflection points. The function is concave up in (, 1/ 2) and (0, 1/ 2); down in ( 1/ 2, 0) and (1/ 2, ). (d) Here is how the graph looks.

4 4 For the functions given in the next two exercises, (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase and find the intervals of decrease. (c) Find the local maximum and minimum values, and state where they occur. (d) Find the intervals where the function is concave up, the intervals where it is concave down, and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph. 8. (Textbook, Section 4.3 # 46) f(x) = x2 4 x We compute the derivatives and set to 0. f (x) = 16x (x 2 = 0, root:x = 0; + 4) 2 f (x) = 16(x2 + 4) 2 2(x 2 + 4)(2x)(16x) (x 2 + 4) 4 = 16(4 3x2 ) (x + 4) 3 = 0, roots: x = 2 3. Applying the second derivative test to the one and only critical point: f (0) = 64/4 3 = 1/4 > 0; there is a relative minimum at 0. We also see that f is positive between its two zeroes (since f (0) > 0); checking f at the left and right of those zeroes, or simply noticing that the sign of f equals the sign of 4 3x 2, and 4 3x 2 < 0 if x is large positive or negative, we see that f is negative to the left of 2/ 3 and to the right of 2/ 3. We are quite ready to answer all the questions. (a) Find the vertical and horizontal asymptotes. The function is defined for all x; there is no vertical asymptote. Since lim f(x) = lim x x x2 4x = 1, lim f(x) = lim x x x2 4x = 1, the line y = 1 is asymptotic at and at +.

5 5 (b) Find the intervals of increase and find the intervals of decrease. Because x = 0 is a local minimum, and there are no other critical points, the function decreases in (, 0), increases in (0, ). (c) Find the local maximum and minimum values, and state where they occur. There is (as mentioned) a local minimum at 0 with value f(0) = 1; there is no local maximum. (d) Find the intervals where the function is concave up, the intervals where it is concave down, and the inflection points. The function is concave down in (, 2/ 3) and in (2/ 3, ); it is concave up in ( 2/ 3, 2/sqrt3). The inflection points are ( 2, f( 2 ) ) 3 3 = ( 2, 1 ) ( 2, 3, f( 2 ) ( 2 ) = 3, 1 ) (e) Use the information from parts (a)-(d) to sketch the graph. For once I ll do this in detail. To get a nice graph we plot all the points we found so far, draw the asymptotes, and then fill in as best we can. Finding the x and y intercepts helps frequently to get a better graph, with the proviso that if the x-intercept is too hard to find, one might just forget about it. In our case, f(0) = 1, so the y-intercept is -1 (and we were already plotting the graph at 0; 0 is critical). The x-intercepts come from solving f(x) = 0; they are 2, 2. We plot the following points: x y To get an even better graph we draw some short horizontal lines at the critical points and short lines having a slope equal to the derivative at that point at the inflection points. This will allow us to modulate better the graph. At the inflection points we make sure the tangent segment goes through the graph. At our inflection points we have ( f ± 2 ) = ± ±0.65 We get the following picture to fill in; the asymptote appears in red. Here is how the graph should look. As you can see the tangent segments at the inflection points have become invisible; they are covered by the curve.

6 6 9. (Similar to textbook, Section 4.3 # 49) f(x) = e x2 /2. As usual we find the critical points and the possible inflection points. We have f (x) = xe x2 /2, zero if and only if x = 0; f (x) = e x2 /2 + x 2 e x2 /2 = (x 2 1)e x2 /2, zero exactly for x = ±1. Because f (0) = 1 < 0, we have a local maximum at 0. We also see from this that f is concave down between 1 and 1. Because x 2 1 > 0 if x < 1 or x > 1, we see that f is concave up to the left of 1 and to the right of 1, showing that we have inflections at ±1. The y intercept is y = 1; there is no x intercept. We are ready to answer the questions. (a) Find the vertical and horizontal asymptotes. There are no vertical asymptotes. Since lim x ± e x2 /2 = 0, the line y = 0 is an asymptote at both and. (b) Find the intervals of increase and find the intervals of decrease. Since we have that single critical point at 0, and the value there is a local maximum, the function increases in ( infty, 0), decreases in (0, ). (c) Find the local maximum and minimum values, and state where they occur. There is a local maximum at 0, of value 1 (also absolute maximum). There is no local minimum. (d) Find the intervals where the function is concave up, the intervals where it is concave down, and the inflection points. The function is concave down in ( 1, 1), up in (, 1), (1, ). The inflection points are ( 1, e 1/2 ) and (1, e 1/2 ). (e) Use the information from parts (a)-(d) to sketch the graph. The table of points to plot is now x y e 1/ Plotting all these points and tangent segments we get the following picture:

7 7 It is now easy to draw a reasonably good plot into this. Here is how it should look. 10. (Textbook, Section 4.3 # 64) As the coffee is being poured in, it is clear that the depth increases slowly at first, then faster as the mug narrows, finally begins to slow again as we pass the midpoint and the mug begins to widen. The graph will look somewhat like The inflection point is the point where the rate at which the depth increases is maximum. 11. Suppose f is a twice differentiable function satisfying f(0) = 0, f(1) = 1 and f(2) = 2.

8 8 (a) Provide two different examples of functions f satisfying the given conditions. (b) Show that for such a function f there is c, 1 < c < 2 such that f (c) = 0. Unfortunately there was a mistake in the statement of what needed to be shown. Part (b) should have said: Show that for such a function f there is c, 0 < c < 2 such that f (c) = 0. (Not 1 < c < 2) Due to this I will consider anything done here bonus points. Part (a) done correctly, 5 bonus points. To get an additional 5 points you have to either figure out what the corrected version of part (b) should have been, and do it. Or simply show that (b) can be false as stated. I might use part (b) again in an exam (maybe in the final exam), so it might be a good idea to figure out how to do it. I would like to emphasize that doing part (b) correctly means coming up with an argument showing that ALL functions f satisfying that f is twice differentiable such that f(0) = 0, f(1) = 1, and f(2) = 2 also satisfy that there is c, 0 < c < 2 with f (c) = 0. For part (a), the easiest function satisfying the conditions is f(x) = x. A second function found by some of you is f(x) = (x 1)