07333000221 2 6 add this value to the set of critical

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Unformatted text preview: s(1) This answer is, of course, the value of f (x0). To see this, first define the function f 1 to be the first derivative of f . Since f is a function, use D. Important: The D operator computes derivatives of functions, while diff computes derivatives of expressions. For more information, refer to the ?operators,D help page. > f1 := D(f); f1 := x → cos(x) esin(x) As expected, f 1(x0) equals the previous limit. > f1(x0); esin(1) cos(1) 7.1 Introductory Calculus • 219 In this case, the second derivative exists. > diff( f(x), x, x ); −sin(x) esin(x) + cos(x)2 esin(x) Define the function f 2 to be the second derivative of f . > f2 := unapply( %, x ); f2 := x → −sin(x) esin(x) + cos(x)2 esin(x) When you plot f and its first and second derivatives, f is increasing whenever f 1 is positive, and f is concave down whenever f 2 is negative. > plot( [f(x), f1(x), f2(x)], x=0..10 ); 2 1 0 –1 –2 2 4 6 8 10 x The graph of f has an inflection point where the second derivative changes sign, and the second derivative can change sign at the values of x where f 2(x) is zero. > sol := { solve( f2(x)=0, x ) }; 220 • Chapter 7: Solving Calculus Problems sol := arctan 2 1√ 1 5− 2 2 √ , −arctan 2 −2 + 2 5 −2 − 2 √ 5), 1√ 1 5− 2 2 √ + π, −2 + 2 5 arctan(− 1√ 5− 2 1√ 5− arctan(− 2 11 , 22 1 1 ,− 2 2 √ −2 − 2 5) Two of these solutions are complex. > evalf( sol ); {0.6662394321, 2.475353222, −1.570796327 + 1.061275064 I, −1.570796327 − 1.061275064 I } In this example, only the real solutions are of interest. Use the select command to select the real constants from the solution set. > infl := select( type, sol, realcons ); infl := arctan 2 > evalf( infl ); 1√ 1 5− 2 2 √ , −arctan 2 −2 + 2 5 1√ 1 5− 2 2 +π √ −2 + 2 5 {0.6662394321, 2.475353222} Observe that f 2 actually does change sign at these x-values. The set of inflection points is given by the following. > { seq( [x, f(x)], x=infl ) }; 7.1 Introductory Calculus • 221 arctan 2 −arctan 2 2 1 1√ √ √ 5− −2+2 5 2 2 , e √ −2 + 2 5 √ 1/2 5−1/2 √ (1/2 5−√ 2)2 1/ 1+4 −2+2 5 , 2 1 1√ √ √ 5− −2+2 5 2 2 + π, e √ −2 + 2 5 √ 1/2 5−1/2 √ (1/2 5−√ 2)2 1/ 1+4 −2+2 5 > evalf( % ); {[0.6662394321, 1.855276958], [2.475353222, 1.855276958]} Since f is periodic, it has, of course, infinitely many inflection points. You can obtain these by shifting the two inflection points above horizontally by integer multiples of 2π . A Taylor Approximation This section illustrates how you can use Maple to analyze the error term in a Taylor approximation. The following is Taylor’s formula. > taylor( f(x), x=a ); f(a) + D(f )(a) (x − a) + 1 1 (2) (D )(f )(a) (x − a)2 + (D(3) )(f )(a) 2 6 1 1 (x − a)3 + (D(4) )(f )(a) (x − a)4 + (D(5) )(f )(a) (x − a)5 + 24 120 O((x − a)6 ) You can use it to find a polynomial approximation of a function f near x = a. > f := x -> exp( sin(x) ); f := x → esin(x) > a := Pi; 222 • Chapter 7: Solving Calculus Problems a := π > taylor( f(x), x=a ); 1 − (x − π ) + 1 1 1 (x − π )2 − (x − π )4 + (x − π )5 + O((x...
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This note was uploaded on 08/27/2012 for the course MATH 1100 taught by Professor Nil during the Spring '12 term at National University of Singapore.

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