Application of DerivativeHard
Question
If f(x) = a ln |x| + bx2 + x has its extremum values at x = - 1 and x = 2, then
Options
A.a = 2, b = - 1
B.a = 2, b = - 1/2
C.a = - 2, b = 1/2
D.none of these
Solution
f′(x) = a/x + 2bx + 1
f¢(- 1) = 0
- a - 2b + 1 = 0
a + 2b = 1
f′(2) = 0
a/2 + 4b + 1 = 0 ⇒ a + 8b + 2 = 0
- 6b = 3 ⇒ b =
, a = 2
f¢(- 1) = 0
- a - 2b + 1 = 0
a + 2b = 1
f′(2) = 0
a/2 + 4b + 1 = 0 ⇒ a + 8b + 2 = 0
- 6b = 3 ⇒ b =
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