MAT 122 Homework 4 Solutions

Transcription

1 MAT 1 Homework 4 Solutions Section.1, Problem Part a: The value of f 0 (1950) is negative. Observe that the tangent line for the graph at that point would appear to be a decreasing linear function, hence have negative slope. This means that the number of farms is decreasing. Part b: The tangent line at t =1960wouldappeartobedecreasingmorequicklythanthetangent line at t = This means that the slope of the tangent line at t =1960ismorenegative, so f 0 (1960) is more negative that f 0 (1980). Section.1, Problem 10 Part a: Agraphofg(t) =(0.8) t is shown below in blue, while the tangent line at t = is graphed in yellow. The slope of this tangent line is the value g 0 (). Since this line has negative slope, the value of g 0 () is negative. Part b: We estimate g 0 () using the small interval [,.01]. The average value of g over that interval is avgroc = g(.01) g() We can therefore estimate that g 0 () = Other answers on this question are possible. We will later show how this derivative can be computed exactly. This will give that g 0 () = ln so we can see that our estimate is relatively accurate. Section.1, Problem 16 Part a: The function appears to be increasing on the entire interval [4, 8], so we can guess that f 0 (6) should be positive. This indicates that the percentage of household with cable TV is growing in the year Part b: To estimate the value of a derivative from a table, we take the average rate of change on the smallest interval possible. Here, these intervals have length. To estimate the value of the derivative at t =, we compute the average rate of change on the intervals [0, ] and [, 4]. These 1

2 are avgroc for [0, ] = avgroc for [, 4] = =1.5 =0.95 Either of these could be an estimate for the value of f 0 (), but we can get another estimate by averaging them, to obtain the estimate f 0 () We repeat the same process for f 0 (10) =1.1 avgroc for [8, 10] = avgroc for [10, 1] = = = Again either of these would be an acceptable estimate, but their average yields another estimate, f 0 (10) =0.375 Any of these estimates has units of % of households/year. Each says that the percentage of people with cable television is growing each year, although we can observe that the growth is slower in the year 000 (t =10)thanintheyear199(t =). Section.1, Problem 18 Point A is a point on the graph of our function, with x-coordinate 4. Since we have that f(4) = 5, the coordinates of point A are (4, 5). Points B and C are points on the line tangent to the graph at x =4. Weknowthatthisline passes through the point (4, 5), and we also know that the slope of this line is 1.5, since f 0 (4) = 1.5. Using point-slope form, the equation of this line is y 5 = 1.5(x 4), y =1.5x +19 Therefore to find the y-coordinates of each point, we need only plug the x-coordinate of each into the equation of this line. Since point C has an x-coordinate of 3.9, it has y-coordinate (1.5)(3.9) + 19 = A similar computation shows that the y-coordinate of B is (1.5)(4.) + 19 = 5.3. Thus the coordinates of B and C are (4., 5.3) and (3.9, 4.85) respectively.

3 Section.1, Problem 0 An approximate reproduction of the graph is shown below, which we ll use to answer the questions. Figure 1: Graph for Problem.1.0 Part a: The value of f(4) is the y-value of the graph of f(x) atx =4. Theyellowdottedline below is the constant function with value equal to f(4). Figure : Graph for Part A Part b: The di erence f(4) f() is the di erence between the y-values of the points on the graph at x = and x =4. Itisrepresentedbythedistancebetweenthetwodottedlinesonthe graph below. Figure 3: Graph for Part B Part c: The value f(5) f() is the slope of the secant line that passes through the graph at 5 x =andx =5,shownbelowonthegraphinyellow. Part d: The value of f 0 (3) is the slope of the line tangent to the graph at the point x =3,shown below on the graph in yellow. 3

4 Figure 4: Graph for Part C Figure 5: Graph for Part D Section., Problem 10 Part a: f(x 3 ) is the largest, as this is the place where the function s value is the largest. Part b: f(x 4 )isthesmallest,asthisistheplacewherethefunction svalueisleast. Part c: f 0 (x 5 )isthelargest,asthisistheplacewheretheslopeofthetangentlinetothegraph is the greatest. Part d: f 0 (x 3 ) is the smallest, as this is the place where the slope of the tangent line to the graph is the least. Notice that this is the only one of the given points where the slope of the tangent line is negative. 4

5 Section., Problem 14 One possible graph is shown below. Notice that the original graph is increasing on the interval ( 1, 1) and decreasing on the interval (1, 1), with a peak at x =1thathasahorizontaltangentline. Thereforethederivativefunctionshouldbepositiveon the interval ( 1, 1), have a horizontal intercept at x =1,andbenegativeontheinterval(1, 1). Figure 6: Graph for Problem 14 Section., Problem 16 One possible graph is shown below. Notice that the original graph in always increasing, so the derivative should be always 0. The original graph also has a horizontal tangent line at x =0,sothederivativefunctionshouldhaveavalueof0atx =0. Figure 7: Graph for Problem 16 Section., Problem 30 Part a: That f 0 (1) > 0 means that the function should have a tangent line at x =1withpositive slope. We see that the two options for this are graphs II and III. In graph III we observe that f 0 (x) is decreasing. As we move right along the graph, the slopes of the tangent lines get smaller, with f 0 () = 0 and negative values for f 0 (x) whenx>. Therefore the correct graph is Graph III. Part b: Again we have that the graph must be either Graph II or III. Since Graph III was the answer to the previous question, we know that the correct graph is Graph II. Part c: By process of elimination, we must have that this is either graph I or IV. Note that this matches with the fact that f 0 (0) < 0, as both graphs I and IV have tangent lines at x =1with negative slope. We observe that as we move right along the graph in Graph IV, the slopes of the tangent lines are decreasing (becoming more negative), so the correct graph is Graph IV. Part d: Process of elimination gives that the correct graph is Graph I. We can also observe, though, that Graph I matches the given description. Section.3, Problem The derivative can be written as dd dt,andhasunitsfeet/minute. 5