Given that the function f (x) = 2lnx-x + a x belongs to [1, e] (where e is the base of natural logarithm), find the function f (x) The known function f (x) = 2lnx-x + a x belongs to the monotone interval of [1, e] (where e is the base of natural logarithm), and the maximum value of known function f (x) is the minimum value of 2ln2

Given that the function f (x) = 2lnx-x + a x belongs to [1, e] (where e is the base of natural logarithm), find the function f (x) The known function f (x) = 2lnx-x + a x belongs to the monotone interval of [1, e] (where e is the base of natural logarithm), and the maximum value of known function f (x) is the minimum value of 2ln2

If f '(x) = 2 / X-1 and f' (x) = 0, then f (x) is monotone in the interval [1,2] [2, e]. If x > 2, f '(x) is less than 0, then f (x) is monotone in the interval [2, e] when X

Let f (x) = 2x power 1 of Xe find the tangent equation 2 of F = (x) at (0, f (0)) and find the monotone interval of F (x)

f'(x)=e^(2x)+2xe^(2x)=(1+2x)e^(2x)
f'(0)=1,f(1)=1
Tangent y = x + 1
Let f '(x) > 0, the solution is x > - 1 / 2
So on (- infinity, - 1 / 2), f '(x)

If the maximum value of F (x) = e − (x − U) 2 is m, and f (x) is even function, then M + U=______ ．

∵ f (x) is an even function, ∵ f (- 1) = f (1), ∵ u = 0 ∵ f (x) = e − X2, ∵ when x = 0, the function f (x) has a maximum value of 1, ∵ m + μ = 1