The function f (x) = ax-a / X-2 * LNX is known. When a = 1, judge whether the function f (x) has extreme value in its definition domain. If so, find it

The function f (x) = ax-a / X-2 * LNX is known. When a = 1, judge whether the function f (x) has extreme value in its definition domain. If so, find it

When a = 1
F(x)=x-1/x-2*lnx
F'(x)=1+1/x ²- 2/x=(1/x-1) ² ≥0
Therefore, f '(x) increases constantly and there is no extreme point

Known function f (x) = 1 + LNX x． (1) If the function is in the interval (a, a + 1) 2) There is an extreme value on, where a > 0, find the value range of real number a; (2) If when x ≥ 1, inequality f (x) ≥ K If x + 1 is constant, find the value range of real number k; (3) Verification: [(n + 1)] 2 > (n + 1) • En-2 (n ∈ n *)

(1) F ′ (x) = - lnxx2, when 0 ＜ x ＜ 1, f ′ (x) ＞ 0, the function f (x) increases monotonically; When 1 ＜ x, f ′ (x) ＜ 0, the function f (x) decreases monotonically. Also, f ′ (1) = 0, ∵ the function f (x) obtains the maximum value when x = 1, ∵ the function has extreme value on the interval (a, a + 12)

Given the function f (x) = the square of lnx-x + ax, obtain the extreme value at x = 1 and find the value of real number a

f(x) = lnx -x^2 +ax
f'(x) = 1/x -2x +a
f'(1) = 1-2+a =0
=> a=1

Known function f (x) = ksinx   The image of passes through point P (π 3， 3) , then the tangent slope of passing point P on the function image is equal to () A. 1 B. 1 two C. - three two D. -1

∵ function f (x) = ksinx   The image of passes through point P (π
3，
3），
∴
3=ksinπ
three
∴k=2
∴f（x）=2sinx  
∴f′（x）=2cosx
∴f′（π
3）=2cosπ
3=1
So choose a

Function f (x) = ㏑ x – ax at point a (2, f (2)), the slope of tangent L is equal to 3 / 2. (1), then find the value of real number a? (2) It is proved that the image of function f (x) is always below the straight line L except point a Request process, thank you!

(1)
∵f(x)=㏑x–ax
∴f'(x)=1/x-a
∴f'(2)=1/2-a=3/2
∴a=-1
(2)
From (1), the linear l equation is y = (3 / 2) x + ln2-1
Construction function H (x) = (3 / 2) x + ln2-1-f (x) = - LNX + (1 / 2) x + ln2-1
H '(x) = - (1 / x) + 1 / 2 (from the question design, x > 0)
∵ when x > 2, H '(x) > 0
When x = 2, H '(x) = 0
x

The slope of the tangent of the function f (x) = x ^ 2 + 2x at point (1,3) is

f'(x)=2x+2
So k = f '(1) = 4