Question

Let f(x) = x 3 _ 3x² a) The interval(s) on which the function is increasing and the intervalls) on which the function f is decreasing B) The relative maximum value of f is and the relative minimum value of f is c) The intervalls) on which the function of is and the intervalls) on concave up which the function F is concare down D) The inflection Point(s) off

Answer 1

Solution ; Given that, fex) = x 3 3x? Now, f(x) = 3x²_6x = 3x(x-2) f(x) = 6X-6 = 6 (x-1) f(x) >o and f(x) <o on (0,0) U (2,0) on (0,2) So, tex) is increasing on (-0,0) U (2,00) and fox) is decreasing on (0,2). b for critical points, f'(x)=0 i 2=0,2 11 At x=0, (0) = -620 At x=2, t f"(2) and attains at x=0 maximum so, relative relative minimum attains at a=2. Then, relative maximum Value of f is fo) = 0 and the relalive minimum value off is fez) = -4.

wow, F"(x) = 6(x-1) 70. on (7,00) and of '(x) <0 on (-2, 1) so the function is concave up on (1,0) and the function fis con cave down (-0,12, 1 on Now, f'(x) = 0 x=t. so, the inflection point of f is a=1.