Absolute maxima and minima a. Find the critical points of f 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 2 ? f(x) = x ? 2x ? 5x + 60n [4,8]

STEP_BY_STEP SOLUTION Step-1 Critical point definition; Let f be a continuous function defined on an open interval containing a 1 number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) 1 = 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 afunction with domain D and let c be a fixed constant in D . Then the output value f( ) 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 a). The given function is f(x) = x 2x 5x + 6 ,0n [4,8] . Clearly the function is a polynomial 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 2x 5x + 6 t hen differentiate the function both sides with respect to x. 1 d 3 2 f (x) = dx ( x 2x 5x + 6) d 3 d 2 d d = dx( x ) dx(2x ) dx5x + dx 6 = 3 x - 2 (2 x) -5 (1) +0 = 3 x - (4 x) -5 since dx (x ) = n x n1 , and dx (Cx) = C dx(x), c is constant. 1 Since , from the definition f (c) = 0 = 3 c - (4 c) -5 2 3 c - (4 c) -5 = 0 (4)+164(3)(5) (4)164(3)(5) Therefore , c = 2(3) , 2(3) = (4)+ 16+60 , (4) 16+60 2(3) 2(3) = (4)+ 76 , (4) 76 2(3) 2(3) = (2+ 19 , (2) 19 , since 76 = 2 19 (3) (3) = 2+4.35889 , (2) 4.35889= 2.1196 , and -0.786296 (3) (3) Therefore , c = 2.1196 , and -0.786296 which is not lie in the interval [4,8] This means the function has no critical points. Therefore , there are no critical points for this function . Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [4,8] so the endpoints are 4 ,8 since from the step -2, the absolute extreme values are ; Therefore , at x = 4 then f(4) = 4 2(4) 5(4) + 6 = 64 - 32-20+6 =18. at x = 8 then f(8) = 8 2(8) 5(8) + 6 = 512 - 128-40+6 = 350. Therefore , f(4) = 18 and f(8) = 350. Hence , the largest value of f(x) is 350 ,this attains at x =8 Therefore , the absolute maximum is f(8) = 350 Hence , the smallest value of f(x) is 18,this attains at x = 4 Therefore , the absolute minimum is f(4) = 18. Step-5 3 2 c). The graph of the related function f(x) = x 2x 5x + 6 ,0n [4,8] is; Hence, from the above graph all the absolute extreme values are true. Therefore absolute values are attains at the endpoints.