Absolute maxima and minima a. Find the critical points off on the given interval. b. Determine the absolute extreme values off on the given interval. c. Use a graphing utility to confirm your conclusions. 3 ?x f(x) = x e , on [-1,5]
Solution 53E Step-1 Critical value definition; Let f be a continuous function defined on an open interval containing a number 1 1 ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 Absolute extreme value definition; When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a fu nction with domain D and let c be a fixed constant in D . hen the output value f (c) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 3 x a). Thegiven function is f(x) = x e on [-1,5] .Clearly the function is a exponential function and it is continuous for all of x . Now , we have to find out the critical points of f on the given interval. Now , f(x) = x e 3 x then differentiate the function both sides with respect to x. 3 dx f(x) = dx(x e x ) 1 3 d x x d 3 d d(v) d(u) f (x) = x dx (e )+ e dx (x ), since dx(uv) = u dx +v dx 3 2 =x ( (1) e x )+ e x/2 (3x ) 3 x x/2 2 3 x 2 x = (-x e )+ e (3x ) = (-x e )+ (3x e ) 2 = x e x ( -x +3) 1 2 c Since , from the definition f (c)=0 = c e ( - c+3) 2 c e c ( - c+3) = 0 c 2 That is , e =0 , c = 0 , and (-c+3) = 0 ln( e c ) =ln(0) , c = 0 , and c=3. (c)ln(e) = undefined , c = 0 , and c= 3. Therefore, c = 0 , 3. Clearly x = 0 , 3 lies between [-1,5]. 3 0 At x = 0, then f(0) = 0 e = 0 . 3 3 At x = 3, then f(3) = 3 e = 27 (0.0498025) = 1.34466 . Therefore ,f(x) attains the critical values at x = 0 3, and the critical points are (0,0) , (3, 1.24466). Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [ -1 ,5] , and the function attains the critical value at x = 0, 3.So the endpoints are -1,5 From the step-2 , the absolute extreme values are; 3 Therefore , at x = 0, then f(0) = 0 e 0 = 0 . 3 At x= -1 , then f(-1) = (1) e (1) = - (e) = -2.718 3 3 At x = 3, then f(3) = 3 e = 27 (0.0498025) = 1.34466 , since, e = 2.718 3 5 At x = 5, then f(5) = 5 e = 125(0.00674144) = 0.842680, since, e = 2.718 Therefore , f(0) =0 , f(-1) = - 2.718 , f(3) = 1.34466 ,and f(5) = 0.842680 Hence , the largest value of f(x) is 1.34466 ,this attains at x= 3 Therefore , the absolute maximum is f( 3) = 1.34466 Hence , the smallest value of f(x) is -2.718,this attains at x = -1 Therefore , the absolute minimum is f( 1 ) = -2.718 Step_5 c) . The graph of the related function f(x) = x e Hence, from the above graph all the extreme values are true.