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. f?? ?) = x (x + 4) ; [?27, 27]

Solution 52E Step-1 Critical point definition ; Let f be a continuous function defined on an open interval containing a number ‘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 values 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 function with domain D and let c be a fixed constant in D. Then 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 1/3 a). The given function is f(x) = x ( x + 4) 0n [ 27,27] .Clearly the function is 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 1/3 ( x + 4) = ( x 4/3 + 4x 1/3) , for the critical points , we have to differentiate the function both sides with respect to x. d d 4/3 1/3 dx f (x) = dx ( x + 4x ) f (x) = d ( x4/3 ) + d (4x1/3 ),since d ( u+v) = d ( u) + d (v) dx dx dx dx dx = ( 3 )( x (4/3)) +4 ( )3 (1/3)1) , since dx ( x ) = nx n1 . 4 1/3 4 2/3 = 3 ( x ) + 3 ( x ) Since , from the definition f (c) = 0 = 4 ( c ) + 4 ( c 2/3)...