Question

2. Sketch the graph of a function where all the following properties hold. For full marks, clearly and carefully label all intercepts, relative extrema, inflection points, and asymptotes. • Domain: (-0,0) • Continuous everywhere • Differentiable everywhere except at x = -3 • f(0) = 6 • lim f(x) = 0 • l'(-2) = f'(0) = 0 • f'(x) < 0 on (-0, -3) and (0,0) • f'(2) >0 on (-3,0) lim f'(x) = 0 and lim f'(x) = -0 000 2-3+ 2-3 f"(x) > 0 on (-2, -1) and (2,00) • f"(x) <0 on (-00,-3), (-3,-2) and (-1,2)

Answer 1

Graph is follows А Inflection point y = f(x) Co, 6) LE F (Inflection point) B 2-3 =-2 X =- x=2 (minima) Asymptotes) Relative maxima = point E = (0,6)

Juslification above graph, obviously (-2, 0). Also, domain is continuous graph is be can also that sein 2= -3. axis is hence g be seen and at x = -2 everywhere. It easily graph smooth everywhere except at x = -3, hence function of will be differentiabl everywhere where except Intercept at f(0) = 6. Also, et can that tangent will horizontal, - 10) = 0 it noticed that function is decreasing for C-00 - 3) Co, o while increasing for justifies that f(a)co (0, 0) while t (21770 for/on (-3,0). be x = 6 therefore f(-2) Also, be can (-3,0) which (-0,3) and on

slightly less where + 2 to Now, at n-3t, tan will make tangent angle than 1/2 le. 7/2-0 Hence, lim f'(a) = D. 23 On other side for n - 3 tangent makes angle slightly greater than Tz and tan (12+) = - hence lim f(n) = -0. 23 Also, it can be noticed that graph is concave downward for (-00, -3) U(-3,-2) concave UE, 2). While upward for (-2,-1) U (20) hence f"(n) so (-2,-1) and (2, 2) while "(2) Lo on (- 0,- 3) and (-3,-2) and (-1,2). at is changing Concauity x = -2, x=-1 and which point of influction - graph is on a=2

is an Also, y=0 hence , ^ , Therefore, asymptore from lim f(x)=0 no conditions have our graph all 01 given been full filled and correct. is