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Let f(x)={4x-x^3+log(a^2-3a+3),0 {x-18 >=3 Then the complete set of 'a' such that f(x)has local maxima at x=3
Dear student f’(x) = 4 – 3x^2 = 1 f(3) – 15 a^2 – 3a + 3 >= 1 (a-1)(a-2) >= 0 a belongs to ( – infinity, 1) U (2, infinity)
Dear studentF’(x) = 4x – 3x^2 + So , the value of x is x>= 15 put x = 15 for this t be local maximum The value of (a^2-3a+3) must be less than 0 So , (a-1)(a-2) must be less than 0 So , value of a must be (-inf , -1) U (1,2) Because there is log and value can not be between (-1 , 1) Hope this helps Good Luck
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