MATH 149LABORATORY ASSIGNMENT 2USING DERIVATIVES TO FIND ABSOLUTE MAXIMA AND MINIMA DERIVATIVES Differentiation is a process that, in most instances, involves only a few rules which are used over and over. Even for relatively simple functions, such as those in the examples and exercises that follow, the results may quickly become rather complicated and unwieldy. Therefore differentiation lends itself very well to execution by a computer. If f has been entered as a function in MAPLE, then the command "D(f);" yields the derivative of f. For example, let f(x) = x2()sec x We will find the first and second derivatives of f.> restart;> f:=x->x^2*sec(x); := f → xx2()sec x> D(f); → x + 2 x ()sec xx2()sec x ()tan x> D(D(f)); → x + + + 2()sec x 4 x ()sec x ()tan xx2()sec x ()tan x2x2()sec x () + 1()tan x2 To compute the n th derivative, we can use "(D@@n)(f);" thus the third derivative of the function f defined above is: > (D@@3)(f);x6()secx()tanx6x()secx()tanx26x()secx() + 1()tanx2x2()secx()tanx3 + + + → 5 x2()sec x ()tan x () + 1()tan x2 + FINDING THE ABSOLUTE MAXIMUM AND MINIMUM The theory tells us that a continuous function defined on a closed interval always has an absolute maximum M and an absolute minimum m; i.e., there are numbers α and β in [a,b] such that m = f(α) < f(x) < f(β) = M for all x in [ a , b ]. Moreover, to find α and β (and thus M and m) we need only consider the endpoints a and b and the critical points, i.e., the solutions to the equation f '(x) = 0, and values of x for which f '(x) does not exist. As an example, we will find the absolute maximum and minimum of f( x ) = sin x + xcos( x2) on the interval [0, π ]. Since we had previously assigned the name "f" to the function x2()sec x , we must enter "restart" if we want to use "f" for a different function. "restart" clears the memory of all previous name assignments made in the current Maple session..> restart;> f:=x->sin(x)+x*cos(x^2); := f → x + ()sin xx()cos x2> plot(f,0..Pi);First use the above graph and the cursor to find approximate values of the absolute maximum and minimum.Next, use the derivative to find exact values of the absolute maximum and minimum: > D(f); → x + − ()cos x ()cos x22 x2()sin x2> plot(D(f),0..Pi,color=green);This function has a derivative at every point. Therefore the only critical points are the solutions of the equation f '(x) = 0.> solve(D(f)(x)=0,x);()RootOf − − + ()cos _Z ()cos _Z22 _Z2()sin _Z2MAPLE cannot find a general solution, so we will use the "fsolve" command to find decimal approximations to the solutions. From the graph, it is clear that there are four solutions of f ' (x) = 0, since the graph of f '(x) cuts the X-axis four times.> X[1]:=fsolve(D(f)(x)=0,x=0.8..1); := X10.9201095708> X[2]:=fsolve(D(f)(x)=0,x=1.6..2); := X21.824276689> X[3]:=fsolve(D(f)(x)=0,x=2.4..2.6); := X32.509682366> X[4]:=fsolve(D(f)(x)=0,x=3..Pi); := X43.086995383 From the graph it is clear that each of the intervals specified in the above four commands contains exactly one zero of f (x). Finally, we calculate the values of f at these four points and at 0 and π, the endpointsof the interval under consideration:> f(X[1]);f(X[2]);f(X[3]);f(X[4]);f(0);f(Pi);1.405270457-0.82463307413.100075094-3.0155005420π ()cos π2For comparison purposes, we calculate a decimal expansion for f( π):> evalf(f(Pi));-2.835869698Therefore the absolute maximum is f (X3) = 3.100075094 and the absolute minimum is f(X4) = -3.015500542. Instead of typing "evalf(f(Pi));" we could have used the percent symbol that acts as a placeholder for the last value computed by Maple. For example, > f(Pi);π ()cos π2Now use the percent symbol: >
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