Question

19. f(x)=4x3- 15x2 18x+ 20 3 (a) Find f'(x) Simplify and factor completely. )12x- 3ox-18 Find f(x) (b) Simplify and factor completely. ( 1.F Intevas are 242-30 (c) Find the intervals where the function is increasing and where it is decreasing. Use interval notation. : 24y-30 O 12 -30x-18O ,3 (-0,). 3). 39 -15 12 3 C 3, 00) 24 -) f is INC on: f is DEC on: Chrek Find the relative extrema of this function if they exist. Identify the types. (d) Find the intervals where the function is concave up and where it is concave down. Use interval notation. (e) fis CU on: f is CD on: Find the coordinates of the points of inflection if they exist. Use appropriate notations. (f)

Answer 1

Then d.co da solution - Here given that fro= 4x3_152²-18x+ 20 @ Now, we calculatef!w. fewaret (923–152²-182+20) = , (923) - (15x?)-$2182) (ise is constant = 122² - 30x-18 70 f'(x) = 12x² – 30x-18 = 6(2x²–5x-3) 261202_62+x-3) = 8 (123+1] (2-3) IF! (20 = 6 (2x+1] (A-3) (and - apow, we calculate f" (2) f'(x)= q f're & I ( 122² - 30 8-18] - $(1222) – 4.BOX) - (18) = 12.2x - 30.1-18.0 = 24x-30 Ifrag z 6(92-5) Any

© now, check whether Hwy Chereasing and decreasing For increasing: f'(x) 70 6(2x+1)(2-3) Yo (2x+1)(x-3) 70 2x+1 70 and x-370 or 2x+Ko ora zlo x>-12 and x>3 or x 2-1/2 or 23 x6 (172900) n (3.00) or KE (-00112) n(-03) XE (300) or at (-00172) at (0o, 1/N (3,00) since few is increasing at xt (-00, (3,000) Por decreasesy foco 6(2x+1) (2-3) co (2x+1)(x-3) Co P (2x + 1) to and (x-3) Lo or (2241) Ko and (x-3) or x 7th and wis or 2x+120 and X-3707 xa-1 RE(-1,00) and af(-0,3) or x € (-00,-/) and xe 13,00 and x>3 7 KG (-22 100) n (-0,3) or at (-00, -1/₂) n (3,000 -(-, 3) 0 1 - 0 | (-2/3) 1 . KE(-1,3 g Since few is decreasing at xe (23) 11 Any

@ 1117777171 Here, we have fay= 4x² - 15x²_18x + 20 Now, need to get derivative so that we can find the critical point for this function on f'(x) = f'(a)= 12x² – 30x-18 to f'1Wzo =) 12 22-302-18 20 26 (2x2 - 5x-390 = 6( 22+D(2-3) 20 x = -1/2 1 3 How to evaluate the funetion at the critical point f(-1/2) = 9 (-1) 3_ 15 (-2/2_18x (-) +20 24x-1 - 15 + 189 +20 = -1-15 + 9+20 82 T - -2-18 + 9 +20 ) - 17+ 116 ११ I fl-2) = 99 . and f(3) = 4 33 – 15 3² -1843 +20 - 9x27 - 15X9-54-720 2 108 - 135-54+20 = -7-54 =-61 1813) = -61 Therefore the free has maximum value and few has minimum value qa at xz - 12 61 ab x=3

© check concave upward or downward we know that function for has concave Upwogodt f'(x) > 0 concave down wond: f'(X) ço fW= 4 x3_158²_182+20 f'(x) = 120²-302-18 f" (V): 243-30 For con care upward fr(2) >0. 242-30 yo x> 30 = 3/5 e ) 269, fulasco 294-30c0 = x <3 x =) X6 (-0,) Porcon came down word : UUUUUUUUUUU since concave upward at concave down ward at af 6 Here given that feu= q 23_15x² 18x720 ' we know that fas has point of inflection of) f (x)=0 f"sal= 294-30 20 = x= 30 - 5 RE y = 40 - 15x2 18x120 at 2:54

20 y = 4 x 2 - 15X52_1345 +20 = 574,-43) - 45 +20 -P51 ) - 45 +20 = 250*5.) - +20 = -25-180+160 = L =43495 the cordenate of point of Inflection (.