Calculus with Analytic Geometry I Exam 4-Take Home Part. Solutions

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1 Calculus with Analytic Geometry I Exam 4-Take Home Part. Solutions 1. (Similar to 2.8#4) The picture shows the graph of a function f. Sketch the graph of f. You may use the grid on the last page, or you can trace the graph on your own paper, and then draw the graph of f underneath it. See the picture on the page before the last page (the penultimate page), showing both curves, with the graph of the derivative in red. 2. (Probably not similar to any text exercise) Suppose f is an even function. Explain as best you can why f will be odd. The graph of an even function is symmetric about the y-axis; the part of the graph on the left side of the y-axis is the reflection of the part on the right side with respect to that axis. That means that the tangent line at x is the reflection with respect to the y-axis of the tangent line at x. The only difference between the line and its reflection is that if you compare rises and runs, the rises will be the same, while the runs will be going in the opposite direction, thus the slope of the reflected line will be minus (the additive inverse of) the slope of the original line. To be slightly more precise, if you have any figure, graph in the plane, if a point of coordinates (x, y) is in the graph, then ( x, y) is in the reflection of the graph about the y-axis. If the graph is a line of equation y = mx+b, the reflected line will have to be y = mx+b. Reflecting a line does not change the y-intercept, but changes the slope from m to m. Returning to graphs of functions and their tangent lines, it means that if f (x) = m, then f ( x) = m; in other words, f ( x) = f (x). We can, of course, be more precise. The derivative at x is defined as being f f(x + h) f(x) (x). Because f is even, f( x) = f(x), h 0 h f( x + h) = f(x h), and if we consider that as h goes to 0, so does h, we get f ( x) = f( x + h) f( x) f(x h) f(x) f(x h) f(x) h 0 h h 0 h h 0 ( h) = f(x h) f(x) = f (x) h 0 ( h) showing that f is odd. While this is very precise, it is an approach lacking intuition. Comment: Many of you answered by stating that the reason was that dxn dx = nxn 1 and if n is even, then n 1 is odd. As it turns out, using mathematics more advanced than the one we have at our disposal right now, one can use this fact to justify the result. But it shows no intuition whatsoever, no feeling for why the derivative of an even function is odd.

2 2 3. (Exercise #46 of Section 2.8.)The figure shows the graph of four functions. One is the position function of a car, one is the velocity of the car, one is its acceleration, and one is its jerk. Identify each curve and explain your choices. Let us assume that the textbook author isn t lying, and that the curves actually represent what they are supposed to represent. One thing we might observe first (and this may or may not be the most obvious thing; how to approach this problem can differ from person to person) is that every curve with one exception develops a negative slope at some point. The derivative of any such curve will have to be negative at those points. The curve without a negative slope is also always positive. The conclusion is that this curve, curve d, cannot be the graph of the derivative of any of the other functions whose graph is given, thus d is the graph of the position function. If we now look at the slope (or slopes of tangent lines) of d, we see that d has horizontal slope at 0, the slope then increases little by little. It is never negative, and of the remaining curves only c is never negative. I have a small problem with the picture. The slopes of tangent lines of d seem to keep increasing, yet c begins to get smaller a little bit before it intersects d. I think the problem is due to a bad reproduction of an original picture in which the curve d actually begins to slowly curve downward. It keeps on increasing, but at a slower and slower rate. However, there being no other curve that does not cross the x-axis, we have to declare that c represents the velocity. For the acceleration we need a curve that is above the x-axis up to close to the point where curves c and d intersect, since c has positive slopes until then, then gets slightly negative slopes. Curve b fits the bill, thus b is the graph of the acceleration function. By eination we can declare a represents the jerk. We should observe, however, that a does indeed behave like the graph of the derivative of the function whose graph is b. For example, it crosses the x axis exactly at the points where the graph of b has a horizontal tangent, etc. 4. (Similar to 2.8,#27) Find the derivative of the function g(x) = 36 x using the definition of derivative. State the domain of the function and the domain of its derivative. 36 (x + h) 36 x 36 x h 36 x g (x) h 0 h h 0 h 36 x h 36 x 36 x h + 36 x h 0 h 36 x h + 36 x (36 x h) (36 x) h 0 h( 36 x h + 36 x) h h 0 h( 36 x h + 36 x) 1 1 = h 0 36 x h + 36 x 2 36 x

3 3 The derivative is g 1 (x) = 2 36 x. The domain of g is (, 36], the domain of g is (, 36). 5. (Not too different from part of # 35 of 3.1) Find the equation of the tangent line to the curve y = x 5 3e x at the point where x = 0. We see that dy dx = 5x4 3e x, evaluating at x = 0 we get the slope of the tangent line, namely m = 3. The value of y when x = 0 is y = 3. Thus the equation of the tangent line is y + 3 = 3x or y = 3x (Exercise #52, 3.1) For what value of x does the graph of f(x) = e x 2x have a horizontal tangent line? This exercise can be rephrased as: For what value of x is f (x) = 0. Now f (x) = e x 2, setting it to 0 we get e x = 2 thus x = ln (Similar to Exercise 75, 3.1) Let f(x) = { x 3 if x 2 mx 2 + b if x 2 Find the values of m and b that make f differentiable everywhere. A somewhat unfortunate typo appears in the statement of the exercise. However, I think there is only one interpretation, and comparing to exercise 75 of 3.1 should have cleared it up. The fact that nobody ed me to ask about it is evidence that the correct interpretation was clear. The function should have appeared as { x 3 if x 2 f(x) = mx 2 + b if x > 2 The function f coincides with a polynomial to the right of 2 and also to the left of 2. So it is differentiable everywhere except possibly at 2. For differentiability, it is necessary (though not at all sufficient) that the function be continuous, so we must have x 2 f(x) x 2 + f(x) = f(2). Now: f(x) x 3 = 8 = f(2); x 2 x 2 f(x) f(x)4m + b, x 2 + x 2 + so a first condition is that 4m + b = 8. The derivative of f, except at x = 2 is given by { f 3x 2 if x < 2 (x) = 2mx if x > 2 (Notice that I am not saying anything about the value at 2). Now, a derivative does not have to be continuous, but it stands to reason that if we set things up so the function f is continuous at 2, then f will be differentiable. There is actually a result that escapes the level of this course that a derivative can be discontinuous, but it cannot have jump discontinuities. Anyway, it stands to reason to equate x 2 f (x) x 2 + f (x). We have x 2 f (x) x 2 3x 2 = 12, and x 2 + f (x) x 2 + 2mx = 4m. We get 12 = 4m, or m = 3. returning to the first equation 4m + b = 8, we now get b = 4. The answer is m = 3, b = 4. The function f is thus defined by f(x) = { x 3 if x 2 3x 2 4 if x > 2 I am attaching as last page a graph generated by Wolframalpha of the function. The function H I am using there is the Heaviside function; H(x 2) is the Heaviside function shifted two units to the right. The p(x) + H(x 2)q(x) will equal p(x) if x < 2, because H(x 2) = 0 if x < 2. If x > 2 it equals p(x) + q(x), because then H(x 2) = 1. To get the function f(x) of the problem I used p(x) = x 3, q(x) = (3x 2 4) x 3.

4 4 The graph shows a little hole at 2; that is probably due to the fact that Wolframalpha s H(x) is not defined for x = 0. It is also possible to check now by the definition that f is differentiable at 2. We have Since f(x) f(2) x 2 x 2 f(x) f(2) x 2 + x 2 we conclude that x 2 f(x) f(2) x 2 x 3 8 x 2 x 2 (x 2)(x 2 + 2x + 4) + 2x + 4) = 12 x 2 x 2 x 2 (x2 3x (x 2)(x + 2) (3x + 2) = 12 x 2 + x 2 x 2 x 2 x 2 f(x) f(2) f(x) f(2) = 12 x 2 x 2 x 2 + x 2 exists (and equals 12); the function is differentiable at 2. NOTE: You ll get full credit for this exercise if you figured out m = 3, b = 4. What I wrote here is only for those who are interested in learning a bit more than what is absolutely necessary. 8. (Exercise 44, 3.2) Suppose that f(2) = 3, g(2) = 4, f (2) = 2, and g (2) = 7. Find h (2) if (a) h(x) = 5f(x) 4g(x). (b) h(x) = f(x)g(x). (c) h(x) = f(x) g(x). (d) h(x) = g(x) 1 + f(x). h (x) = 5f (x) 4g (x); h (2) = 38. h (x) = f (x)g(x) + f(x)g (x); h (2) = 29. h (x) = f (x)g(x) f(x)g (x) g(x) 2 ; h (2) = h (x) = g (x)(1 + f(x)) g(x)f (x) (1 + f(x)) 2 ; h (2) = (Exercise 48, 3.2) If f(2) = 10 and f (x) = x 2 f(x) for all x, find f (2). We have f (x) = d dx (x2 f(x)) = 2xf (x) + x 2 f(x) = (2x 3 + x 2 )f(x). Evaluating for x = 2 we get f (2) = (Partially Similar to Exercise #35, 3.3) A mass on a spring vibrates horizontally on a smooth level surface. Its equation of motion is x(t) = 36 cos t, where t is in seconds and x is in centimeters. In other words, the mass is moving back and forth on a straight horizontal line that is taken as the x-axis, and the formula gives its position at all times. (a) Find the velocity and acceleration at time t. The velocity is v(t) = x (t) = 36 sin t, the acceleration, a(t) = v (t) = 36 cos t.

5 5 (b) What is the first value of positive t for which the velocity is 0? 36 sin t = 0 for t = π. That is the first positive value at which the velocity is 0. (c) Find the position, velocity, and acceleration of the mass at time t = 2π/3. In what direction is it moving at that time? (Just in case: The directions are left and right.) Setting t = 2π/3 in the formulas we are given for the position, and those we obtained for the velocity and acceleration, we see that the position is x = 18 cm, the velocity v = 18 3 cm/s, and the acceleration a = 18 cm/s 2. Since the velocity is negative, it is moving to the left. 11. (Exercise # 50, 3.3) Find d35 (x sin x) by finding the first few derivatives and observing the pattern that dx35 occurs. d 35 (x sin x) = x cos x 35 sin x. dx35

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