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. ? ? ? f(? ? ) = si? cos x? on [0, 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 a). The given equation is ; f(x) = sin(x) cos(x) , here f(x) is trigonometric function and it is continuous for all of x.The given interval is [0, 2] Now , f(x) = sin(x) cos(x) then differentiate the function both sides with respect to x. 1 d f (x) = dx(sin(x)cos(x)) = sin(x) d cos(x) +cos(x) d sin(x), since d (uv) = u d(v)+v d(u) dx dx dx dx dx d d = sin(x)(-sin(x))+cos(x)cos(x) ,since dx sin(x) = cos(x), dx (cos)=-sin(x) = cos (x) sin (x) = cos(2x) . Since , from the definition f (c)=0 = c os(2x) . That is, cos(2x) = 0. cos(2x) = cos( ),2because cos( ) = 0 2 2x = 2 x = 4 . At , x = 4 then f(x) = sin( 4cos( ) 4 =( 1 )( 1 ) 2 2 = (½) Therefore , f(x) attains the critical value at x= , and the endpoints of the interval is x= 4 0,2 . Hence, the critical point is ( , ) 4 2 Step_3 b). The graph of the related function is : f(x) = sin(x)cos(x) , on[0,2] Therefore, from the graph it is clear that the function attains neither local maximum nor local minimum at x= . 4