The monotone decreasing interval of the function y = half power INX is?

The monotone decreasing interval of the function y = half power INX is?

Decreasing then y '

The monotone decreasing interval of function y = x-inx, X belonging to (0, positive infinity) is

y=x-lnx
y'=1-1/x=0,x=1
y''=1/x²>0
The monotone decreasing interval is: (0,1)
The monotone increasing interval is: (1, + ∞)

How to find the monotone increasing and decreasing interval of function INX / X

F (x) = LNX / X f '(x) = (1-lnx) / x ^ 2 = 0, x = e monotonically increasing (0, e), decreasing interval (E, positive infinity)

The function f (x) = inx-2x is known
seek
(1) Find the monotone interval of function f (x);
(2) Finding the tangent equation of function f (x) at point (1. F (x))

(1) Let f (x) ′ = 1 / X-2, Let f (x) ′ = 0 get x = 1 / 2, that is, the domain of definition of stationary point original function is (0, + ∝), so the monotone interval is: (0,1 / 2), (1 / 2, + ∝) 1 / 2 is a point with zero derivative and has extremum; (2) substitute x = 1 into f (x) ′ = 1 / X-2 to get f (x) ′ = - 1, that is, tangent slope - 1

Given that the function f (x) = 2aX ^ 3-bx "- 6x has an extreme value at x = - 1 or x = 1, try to find the tangent equation of function f (x) at x = - 2?

The derivative of F (x) = 6AX ^ 2-2bx-6, substituting x = - 1 and x = 1 into the order = 0, we get a = 1, B = 0, f (x) = 2x ^ 3-6x, f (x) derivative = 6x ^ 2-6, f (- 2) = - 4 when x = - 2, f (x) derivative = 18 when x = - 2, which is the slope tangent of tangent equation, k = 18
Passing point (- 2, - 4) y = 18x + 32

Function f (x) = ax-b / x + INX when f (1) = 0, if function f (x) is a monotone function, find the value range of real number a
When f (x) reaches the extremum at x = 2 and x = 4, the equation f (x) has three different real roots in the interval [1,8]. The value range of real number C is obtained

Because: F '(x) = a + B / X & # 178; + 1 / X
Because f (x) is a monotone function, f '(x) > 0 or F' (x) < 0
And: F (1) = A-B = 0, a = B
So: F '(x) = a + B / X & # 178; + 1 / x = a + A / x2 + 1 / X
Let t = 1 / x, then f ′ (x) = at & # 178; + T + A. (a ≠ 0)
If f ′ (x) < 0, then a < 0, and 1-4a & # 178; < 0, then a < - 1 / 2
If f '(x) > 0, then a > 0, and 1-4a & # 178; < 0, then a > 1 / 2
If a = 0, then f '(x) = 1 / X ＞ 0, so the condition is also satisfied
So: a ＜ - 1 / 2, a ＞ 1 / 2 or a = 0

Let g (x) = f (x) / x, and find the maximum value of G (x)

If a = 1. F (a) = - 2. A = 2. F (x) = x2-4x + 2 g (x), the minimum value is 2 times root 2 minus 4

On the image of function y = LNX, the abscissa of the nearest point to the line 2x-y + 2 = 0 is the conjugate complex z = √ 2 + 1 / 2 - √ 2I?

The closest point from y = lnxy '= 1 / X to the line 2x-y + 2 = 0 is the tangent point parallel to the line. Then the slope of the tangent k = 2, i.e. 1 / x = 2, gets x = 1 / 2Y = ln1 / 2 = - LN2, i.e. the point coordinate is (1 / 2, - LN2) 2. Z = (1 - √ 2I) / (√ 2 + I) = (1 - √ 2I) (√ 2-I) / [(√ 2 + I) (√ 2-I)] = (- 3I)

The general picture of the function f (x) = lnx-12x2 is ()
A. B. C. D.

∵ f (x) = lnx-12x2, its definition domain is (0, + ∞) ∵ f ′ (x) = 1x-x = (1 − x) (1 + x) x, from F ′ (x) > 0, 0 < x < 1; F ′ (x) < 0, x > 1; ∵ f (x) = lnx-12x2, monotonically increasing on (0,1), monotonically decreasing on (1, + ∞); ∵ when x = 1, f (x) takes

If the function f (x) = MX ^ 2 + lnx-2x is an increasing function in the definition and, then the value range of M

The domain is (0, positive infinity)
Derivation f '(x) = 2mx + 1 / X-2 = (2mx ^ 2-2x + 1) / X
If x > 0, f '(x) > = 0
So any x > 0,2mx ^ 2-2x + 1 > = 0, M > 0
The axis of symmetry of the function g = 2mx ^ 2-2x + 1 is x = 1 / 2m > 0
Therefore, m needs to satisfy △ = 4-8M = 1 / 2
To sum up, M > = 1 / 2