Given the function f (x) = LNX / x, if a > 0, the function H (x) = XF (x) - x-ax ^ 2 has an extreme value on (0,2), find the value range of real number a

Given the function f (x) = LNX / x, if a > 0, the function H (x) = XF (x) - x-ax ^ 2 has an extreme value on (0,2), find the value range of real number a

h(x)=lnx-x-ax ²
h'(x)=1/x-1-2ax=-(2ax ²+ x-1)/x
If there is an extreme value, then 2aX ²+ X-1 = 0 has a single root in (0,2) (cannot be a double root, i.e. a ≠ - 1 / 8)
By a = (1-x) / (2x) ²)= 1/2[1/x ²- 1/x]=1/2(1/x-1/2) ²- 1/8
Because 1 / x > 1 / 2, there is a > - 1 / 8
That is, the value range of a is a > - 1 / 8

Let the function f (x) = LNX + A / (x-1) have extreme values in (0,1 / E): (1) find the value range of real number a; (2) If x1 ∈ (0,1), X2 ∈ (1, + ∞). Verify that f (x2) - f (x1) > e + 2-1 / E The first question has been made. Let's look at the second question, that is, to verify: F (x2) - f (x1) > e + 2 - (1 / E.)

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Let the function f (x) = LNX + A / (x-1) have extreme value in (0,1 / E) (1) find the value range of real number a (2) if x1 ∈ (0,1), Let the function f (x) = LNX + A / (x-1) have an extreme value in (0,1 / E) (1) Find the value range of real number a (2) If x1 ∈ (0,1), X2 ∈ (1, + ∞), verify that f (x1) - f (x2) > e + 2-1 / E. (E is the base of natural logarithm)

a

Known function f (x) = 1 2(x−1)2+㏑x−ax+a． (1) If a = 3 2. Find the extreme value of function f (x); (2) If f (x) > 0 holds for any x ∈ (1,3), find the value range of A

(1) The definition field of function f (x) is (0, + ∞). F '(x) = x − 1 + 1x − A. when a = 32, f' (x) = x + 1x − 52 = 2x2 − 5x + 22x, Let f '(x) = 0, and the solution is x = 12 or 2. List: X (0, 12) 12 (12, 2) 2 (2, + ∞) f' (x) + 0 -

Find the extreme value of function f (x) = LNX / X

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If the function f (x) = ax ^ 2 + 2x + b * LNX takes the extreme value at x = 1 and x = 2 (1) Find the value of a and B (2) Find the maximum and minimum values on [1 / 2,2]

Derivation of F (x)
f'(x)=2ax+2+b/x
Take the extreme value of x = 1 and x = 2. Obviously, substitute f '(x) = 2aX + 2 + B / x, i.e. 0
2a+2+b=0
4a+2+b/2=0
Simultaneous, solution, a = - 1 / 3, B = - 4 / 3
2)f(x)=-1/3x^2+2x-4/3*lnx
f(1/2)=11/12+4/3ln2
f(1)=5/3
f(2)=8/3-4/3ln2
Therefore, the minimum value is 5 / 3 and the maximum value is 11 / 12 + 4 / 3ln2