bạn sẽ cần phải kiểm soát phạm vi nâng cấp và thay đổi. Trong Rose, bạn có thể xem trường hợp những người tham gia sử dụng bằng cách sử dụng menu Báo cáo. Để xem các lớp học và các hoạt động tham gia trong một trường hợp sử dụng: 1. Chọn các trường hợp sử dụng trên một sơ đồ tình huống sử dụng. | CHAPTER 6 Symbolic Calculus Differential Eqs 155 J EXAMPLE 6-4 Does y t 3 sin t 7 cos 5t solve y y -5 cos 2t SOLUTION 6-4 Define our function y 3 sin t 7 cos 5 t Now let s create a variable to hold the required result f -5 cos 2 t To enter the left-hand side of the differential equation we create another variable a diff y 2 y We use isequal to check whether the equation is satisfied isequal a f ans 0 Since 0 is returned y t 3 sin t 7 cos 5t does not solve y y -5 cos 2t. EXAMPLE 6-5 Find the minima and maxima of the function f x x3 - 3x2 3x in the interval 0 2 . SOLUTION 6-5 First let s enter the function and plot it over the given interval syms x f x 3-3 x 2 3 x ezplot f 0 2 The plot is shown in Figure 6-4. To find local maxima and minima we will compute the derivative and find the points at which it vanishes. The derivative is g diff f g 3 x 2-6 x 3 Figure 6-4 A plot of f x - x3 - 3x2 3x As a quick aside we can use the pretty command to make our expressions look nicer pretty g 2 3 x - 6 x 3 Well slightly nicer anyway. Returning to the problem let s set the derivative equal to zero and find the roots s solve g s 1 1 We see that there is only one critical point since the derivative has a double root. We can see from the plot that the maximum occurs at the endpoint but let s prove this by evaluating the function at the critical points x - 0 1 2. We can substitute a value in a symbolic function by using the subs command. With a single variable this is pretty simple. If we want to set x - c we make the call CHAPTER 6 Symbolic Calculus Differential Eqs 157 J subs fc . So let s check f for x 0 1 2. We can check all three on a single line and have MATLAB report the output by passing a comma-delimited list subs f 0 subs f 1 subs f 2 ans 0 ans 1 ans 2 Since f 2 returns the largest value we conclude that the maximum occurs at x 0. For fun let s evaluate the derivative at these three points and plot it subs g 0 subs g 1 subs g 2 ans 3 ans 0 ans 3 Where are the critical points
