Let f (x) = ax LNX (a is a constant) (1) find the maximum value of function FX when a = 1. (2) discuss the maximum value of function FX at (0, ∞)

Let f (x) = ax LNX (a is a constant) (1) find the maximum value of function FX when a = 1. (2) discuss the maximum value of function FX at (0, ∞)

If f (x) = x-lnx, X belongs to (0, + ∞) f '(x) = 1-1 / X. Let f' (x) = 0, the solution x = 1 (0,1) decrease, (1, + ∞) increase x = 1, there is a minimum value f (1) = 1 LIM (x tends to 0) f (x) = + ∞ LIM (x tends to + ∞) f (x) = + ∞ so the minimum value is 1, and there is no maximum value. (2) f '(x) = A-1 / x, X belongs to (0, + ∞) when a ≤ 0, f

The monotone increasing interval of the function f (x) = LNX ax (a > 0) is______ .

The domain of ∵ f (x) is (0, + ∞),
Then f ′ (x) = 1
x-a，
Let f ′ (x) > 0, the solution is 0 ＜ x ＜ 1
a．
So the answer is: (0, 1
a）

Let f (x) = ax - (a + 1) LNX, where a ≥ - 1, find the monotone interval of F (x)

First, x > 0
f'(x)=a-(a+1)/x
Let f '(x) = 0 get x = (a + 1) / a by x > 0 a > = - 1
a> When 0, x = (a + 1) / a satisfies f '(x) = 0
When 00, the function increases in this interval
-1

Given a > 0, the function f (x) = LNX ax ^ 2 (x > 0), find the monotone interval of F (x)

f′(x)=1/x-2ax
1/x-2ax=0
1/x=2ax
X = ± radical (1 / 2a)
Because x > 0, x = under root sign (1 / 2a)
When x = 0, the function increases monotonically
When x > = root (1 / 2a), f '(x)

Let f (x) = LNX, G (x) = A / x, Let f (x) = f (x) + G (x). (1) when a = 1, find the monotone interval of function f (x) (2) If the tangent slope k ≤ 1 / 2 of any point P (x0, Y0) on the image of function y = f (x) (0 < x ≤ 3) holds, the minimum value of a is obtained

1 f (x) = ln x + 1 / X (x > 0) derivative f '(x) = 1 / X-1 / (x ^ 2)
When f '(x) = 0, x = 1
Zero

Given the function f (x) = ax ^ 2 - (a + 2) x + LNX (Ⅱ), find the monotone interval of function f (x) Please write a complete process, the best picture!

The domain of F (x) is x > 0
f`(x)=2ax+(a+2)+1/x
=(2ax^2+(a+2)x+1)/x
=(ax+1)(2x+1)/x
When a > = 0
f`(x)>0
F (x) increases monotonically on (0, + ∞)
When a < 0
Let f '(x) > = 0
X < = - 1 / 2 or x > = - 1 / A
∵x>0
﹣ f (x) is reduced on (0, - 1 / a)
F (x) increases monotonically on [1 / A, + ∞)

Function f (x) = LNX / X (x > o) monotone decreasing interval

Find the monotone decreasing interval of function f (x) = LNX / X
The function f (x) = LNX / X is defined as x > 0
f'(x)=[(1/x)*x-lnx*1]/x^2=(1-lnx)/x^2
Then, when 1-lnx < 0, i.e. LNX > 1, i.e. x > e, f '(x) > 0
Therefore, the increasing interval of the function f (x) = LNX / X is x ∈ (E, + ∞)

Given that f (x) = LNX of X, what is its monotone decreasing interval?

f'(x)=(1/x*x-lnx*1)/x²
=(1-lnx)/x²
Then f '(x) 0
So 1-lnx1
X>e
So the minus interval is (E, + ∞)

The monotone decreasing interval of the function f (x) = x Λ 2 / 2-lnx is

∵f'(x)=(x+1)*(x-1)/x (x>0)
Let f '(x)

Given the function f (x) = x ^ 2 + ax LNX, a belongs to R. (1) if the function f (x) is a decreasing function on [1,2], find the value range of real number A. (2) Let G (x) = f (x) - x ^ 2 Which is the college entrance examination question?

Given the function f (x) = x ^ 2 + ax LNX, a  R. ① if the function f (x) is a decreasing function on [1,2], find the value range of the real number a. ② let g (x) = f (x) - x ^ 2, if x  0, e], the minimum value of G (x) is 3, and find the value of A
1. f'(x) = 2x+a+(-1/x)
=>When x belongs to [1,2], f '(x) is an increasing function
=> f'(1) a+1 g'(x)=a-(1/x)
=>When x belongs to (0, e], G '(x) is an increasing function
=>G '(x) g (x) > = g (E) = AE-1, the minimum value of function g (x) is 3
=> ae-1 = 3
=> a = 4/e
(2) When (x) is (x), G is a function
When x ∈ (1 / A, e] = > G (x) is an increasing function
=> g(x)>=g(1/a)=1-ln(1/a) =3
=> ln(1/a) = -2
=> a = e^2