Locating critical points a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing unity In determine whether each critical point corresponds to a local minimum. local maximum. or neither. 5 3 f(x) = 12 x - 20 x , [-0.5,2]

STEP_BY_STEP SOLUTION Step-1 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) = 0 1 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). 4 2 Example ; Find all critical points of f(x) = x 8x . Because f (x) is a polynomial function, its domain is all real numbers. 1 3 f (x) = 4x 16x f (x) = 0 =4x 16x 2 4x ( x - 4) = 0 2 4x(x+2)(x-2) = 0 , since a - b = (a+b)(a-b) X = 0 , x = -2 and x=2. 4 2 At x= -2 , then f(-2) = ( 2) 8( 2) = 16 - 32 = -16. At x= 0 , then f(0) = (0) 8(0) = 0 - 0 = 0. 4 2 At x= 2 , then f(2) = (2) 8(2) = 16 - 32 = -16. Hence, the critical points of f(x) are (2,16), (0,0), and (2,16). Step-2 5 3 a). The given equation is ; f(x) = 12 x - 20 x , here f(x) is a polynomial function , its domain is all real numbers and the given interval is [-0.5,2] so, f(x) is continuous for all of x. 5 3 Now , f(x) = 12 x - 20 x , then differentiate the function both sides with respect to x. 1 4 2 d n n1 So , f (x) = 60x - 60 x , since dx (x ) = n x Since from the definition f (c) = 0 = 60c - 60 c4 2 2 2 60c ( c 1) = 0. That is , c =0 , and c2 = 1 c= -1, 1. Therefore , f(x) attains the critical values at x = 0,-1 and 1.The given endpoints are x = -0.5 and2. Therefore , by using the number line we can easily say that f(x) attains the local maximum at x=1, because at x=1 the function is increasing .So, from the above the function has no local maximum.That is the local minimum attains at the critical point x=1. Step-3 5 3 b). The graph of the related function f(x) = 12 x - 20 x is;