T, ;,~ .. r': ~\ . . "~ •• '!:I .:' i 1}' J . {:; " . ' " 'i :.-, ,.' ~:. '. ", ,.' .' , . '. MATH 5 MODULE 4 (2HW4) ~TERMED~TEVALUETHEOREM INSTRUCTIONS: Do not write anything on this page. List your answers a~r PROBLEM 4. FiRin the missing inbrmation in numbers with parentheses ill· PROBlEM1. Verify IVT given the function fdefinedby f(x)=4+3x-x 2 in the il~rval 2s X s 5 fork = 1. SOlUTION1. To verify the in~rmediate vakJe theorem ifk = 1 we need to find a nurroer c i1 the intelVall.{1}, WI such that f(c) = ro. Because f is a polynomial function, it is continuous on its domail Mi. Henoe, it is continuous on the closed intelVal [2,5) Sinre f(2) = m =t:- f(5) = lID , IVT guarantees that there is a nurrber c between m and @ such that f(c) = 1. That is, 3±J21 f(c)=4+3c-C 2 =1 q c 2 -3c-3=O q c 2 However, m is an extraneous oolution because this number is outside the interval [2, 5]. Therefore, we accept only the number 1m wh<h is in lhe desired illeJValand f( 3 +f21 ) =1 I t I I PROBlEM2. Verify IVT given f(x)=~ 25 _X2 il the interval -4.5 s X s 3 for k = 3. SOlUTlON2. To verify the intermediate vakJe theorem ifk = tt1l we need to find a number c in the intelVal [-4.5, 3) such thatf(c) = 3. The function f is continuous on its domain M 01ll. Sinoo f(-4.5) = tl.4l =t:- f(3) = @, IVT guarantees that flare is a nurroerc between -4.5 and 3 sum that f(c) = (OO. That is, f(C)=~ 25-c 2 =3 q 9=25-c 2 q c=±4 However, tl1l is an extraneous oolution because this number is outside the interval [-4.5, 3]. Therefore, we acoept only the number {!§l which is in the desired interval and f(-4 ) = @. r f: I, f: >, 4 PROBlEM3. Verify IVT given the function fdefined by f(x)=-- in the in1erva1 -3 s X s 1for k = ~. x+2 SOlUTION3. To verify the intermediate vakJe theorem if k = ~ we need b find a number c in the interval [(20), @J such that f(c) = tm. Because f is a rational function, it is continuous on its domain @. Henoe, it is discontinuous on the interval [-3, 1). Thus, NT cannot guarantee the existenoe of a c between -3 and 1 so that f(c) = ~. 4 1 ToiDustratefurther, f(-3)={H} =t:- f(1)[email protected], f(c)=--=- q c=6 c+2 2 But then C = 6 (2: [-3,1].REMEMBER, WE CANNOT USE IVT WHEN THE FUNCTION IS DISCONTINUOUS!!! , I " fdlmgutielTez I I' I·