The minimum value of function f (x) = 12x2 − LNX is______ ．

The minimum value of function f (x) = 12x2 − LNX is______ ．

∵ function f (x) = 12x2 − LNX  f ′ (x) = x & nbsp; − 1x (x > 0) Let f ′ (x) = x & nbsp; When x ∈ (0,1), f ′ (x) < 0, when x ∈ (1, + ∞), f ′ (x) > 0, so in the interval (0,1), the function f (x) is a decreasing function, in the interval (1, + ∞), the function f (x) is an increasing function, then when x = 1, the minimum value of the function is 12, so the answer is: 12

Given the function f (x) = 1 / 2x Λ 2 + LNX, find the maximum and minimum value of function f (x) in the interval [1, e]

f'>0
f min= f(1)=1/2
f max=f(e)=1+1/2e^2

The minimum value of function f (x) = 12x2 − LNX is______ ．

∵ function f (x) = 12x2 − LNX  f ′ (x) = x & nbsp; − 1x (x ＞ 0) Let f ′ (x) = x & nbsp; − 1x = 0, and the solution is x = 1 ∵ when x ∈ (0,1), f ′ (x) ＜ 0, when x ∈ (1, + ∞), f ′ (x) ＞ 0, so in the interval (0,1), function f (x) is a decreasing function, in the interval (1, + ∞)