Question

Question 7 Given: f(t) = 23 – 672 - 9x + 2. Find the ABSOLUTE MINIMUM value off (z) on (-1,2). 2 - 4 . -6 - - 14 : none B

Answer 1

Son Given x² - 6x - x + 2 ... (1) f(x) to find! Absolute MINIMUM value of f(x) on [1, 6] Now, we will derivate the function f(x) with respect to a f'(x) = 3x² - 12x - 9 Now we put f'(x) =0 to get to get critical points. 3x²_12x-g=0 32 (in whole ear {dividing by x² - 4x - 3 x = - b ± √b2-4 ac qa above quadratic eqh gives a=1, b = -A & C = -3 x = -(-4) ± √(-4) - 4x1x(-3) 2) x = 4 + 16+12 J28 x = u t = 4 2 257 2 3 226 2 uc We get two critical pointe, x=2+57. & Not in a = 2-57 {alot in interually

coitical point interval 1 Now we have to find critical point on the we get one .. u=2-57 Hence leatulating f(x) at a = -1, 2-57, 5 Given f(x) = x 3 - 6x² - gx + 2 f(1) = (-1266-79(-1)+2 -|-6 + 9 +2 (1-7 = 4 $(2-57) = (2-53)-662-54)?- 9(2+56) +2 6 5.09 1) f (2) (239-6(e) - 9(2) + 1/24 - 18 t 2 13 8 - 42 tlo - 32 is -32 at x = 2 Absolute Minimum value none option (d) is correct