Given that f (x) = - x 2 + ax + 1-lnx, if f (x) is a decreasing function on (0,1 / 2), find the range of real number a

Given that f (x) = - x 2 + ax + 1-lnx, if f (x) is a decreasing function on (0,1 / 2), find the range of real number a

Axis of symmetry = - 2A / b = A / 2
Because it's a minus function on 0 to 1 / 2 and the opening is down
So you can draw a picture
If the axis of symmetry is exactly equal to 0, then it is in accordance with the meaning of the question
If the axis of symmetry is greater than 0, then there may be an increase, a decrease, or even an increase between 0 and 1 / 2
If the axis of symmetry is less than 0, it can be guaranteed that it must be a minus function between 0 and 1 / 2
So to sum up
Axis of symmetry

Let f (x) = ax ^ 2-lnx 1. If f (x) gets the extremum at x = 1 / 2, find the value of real number A. if f (x) is a minus function on interval 1,2, find the range of A

(1)
f'(x)=2ax-1/x
F (x) reaches the extreme value at x = 1 / 2
f'(1/2)=a-2=0
Then a = 2
（2）
F (x) is a decreasing function on [1,2],
Then x ∈ [1,2], f '(x) ≤ 0 holds
That is, 2aX ≤ 1 / X
That is, 2a ≤ 1 / x 2 is always true
∵1/x²∈[1/4,1]
∴2a≤1/4
∴a≤1/8

Let x + X be a real number over the range of [1,2] = (2,2)

g'(x)=2x+a/x-2/(x^2)
g"(x)=2-ax^(-2)+4x^(-3)
g'(1)=a

Let f (x) = LNX ax (a ∈ R, a > 0) (1) Find the monotone interval of function f (x); (2) Find the minimum value of function f (x) on [1,2]

(1) The definition domain of function f (x) is (0, + ∞)
f′（x）=1
x-a=1−ax
X (2 points)
Because a ＞ 0, Let f ′ (x) = 1
X-a = 0, x = 1
a；
When 0 < x < 1
When a, f ′ (x) = 1 − ax
X > 0; when x > 1
When a, f ′ (x) = 1 − ax
x＜0，
So the monotone increasing interval of function f (x) is (0,1)
a) The monotone decreasing interval is (1
a. (4 points)
(2) When 0 < 1
When a ≤ 1, i.e. a ≥ 1, the function f (x) is a decreasing function on the interval [1,2],
The minimum value of F (x) is f (2) = ln2-2a. (6 points)
② When 1
I.e. a ≤ 0, a ≥ 1
When 2, the function f (x) is an increasing function on the interval [1,2],
The minimum value of F (x) is f (1) = - A. (8 points)
③ When 1 < 1
A < 2, i.e. 1
When 2 < a < 1, the function f (x) is at (1,1)
a) It is an increasing function on (1)
a. 2) is a minus function
And ∵ f (2) - f (1) = ln2-a,
When 1
When 2 < a < LN 2, the minimum value of F (x) is f (1) = - A;
When LN2 ≤ a < 1, the minimum value of F (x) is f (2) = ln2-2a. (10 points)
In conclusion, when 0 < a < LN 2, the minimum value of function f (x) is f (x) min = - A;
When a ≥ LN2, the minimum value of function f (x) is f (x) min = ln2-2a. (12 points)

Let f (x) = LNX + (1-x) / ax, where a is a constant greater than zero. (2) find the minimum value of function f (x) on the interval [1, e]

f(x)=lnx+(1-x)/ax
=LNX + 1 / AX-1 / a derivation
F '(x) = 1 / X-1 / (AX ^ 2). When f' (x) = 0, that is, x = 1 / A, the function f (x) has an extreme value
So when 1 ≤ 1 / a ≤ e, i.e. 1 / E ≤ a ≤ 1, minf (x) = f (1 / a) = 1-1 / a-lna
When a < 1 / E, minf (x) = f (E) = (ae-e + 1) / AE
When a > 1, minf (x) = f (1) = 0

The increasing function LNY = LNY is monotone

Any value x2 > X1 > 0 x2 / X1 > 1
lnx2-lnx1=ln(x2/x1)>0
So it is incremented over the definition field (0, positive infinity)

The monotone increasing interval of the function y = LNX + e ^ x is

Derivative, y '= 1 / x + e ^ x
Because when the definition domain is x > 0, and y '> 0, x > 0, therefore
The monotone increasing interval of the function y = LNX + e ^ x is
{X|x>0}

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

f'(x)=1－(1/x)=(x－1)/(x)
Then: F (x) is 0

Function f (x) = x The monotone decreasing interval of LNX is______ .

It is known that f ′ (x) = LNX − 1
ln2x，
When 0 ＜ x ＜ E and X ≠ 1, f ′ (x) ＜ 0,
So the function f (x) = X
The monotone decreasing interval of LNX is (0,1), (1, e)
So the answer is (0,1), (1, e)

Monotone decreasing interval of function f (x) = x / LNX

First we get x ^ 2 / (lnx-1), and the monotone decreasing interval is that the derivative is negative, that is (0, e)