Application of DerivativeHard
Question
If f(x) = tan-1x - (1/2) ln x. Then
Options
A.the greatest value of f(x) on [1/ √3, √3] is π/6 + (1/4) ln 3
B.the least value of f(x) on [1/ √3, √3] is π/3 - (1/4) ln 3
C.f(x) decreases on (0, ∞)
D.f(x) increases on (-∞, 0)
Solution
f′(x) = 
, x > 0
=
≤ 0 ∀ x > 0
= f(x) is decreasing ∀ x > 0.
On
, greatest value is
and least value is
f(√3) =
ln√3
, x > 0=
≤ 0 ∀ x > 0= f(x) is decreasing ∀ x > 0.
On
, greatest value is and least value isf(√3) =
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