Vikas TU
Last Activity: 7 Years ago
Dear Student,
f(x) = 5 + (a-2)x + (a-1)x2 – x3.
The extremum occurs when f'(x) = (a-2) + 2(a-1)x – 3x2 = 0.
For x1, the condition for extremum is 3x12 - 2(a-1)x1 – (a-2) = 0.
For this x1 to be minimum, f''(x1) gretare than equals 0.
6x
1 – 2(a-1) ≥ 0 => 6x
1 ≥ 2(a-1) => 3x
1 ≥ a-1 => 3x
1 + 1 ≥ a.
Henc ethe possible values of a would be in term of x =>
a less than equalls to all the values 3x
1 + 1.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
