Monotonically increasing interval of function f (x) = 2x square - LNX A (0,1 / 2) B (0,4 radical 2) C (1 / 2, + infinite) d (- 1 / 2,0) and (0,1 / 2) what should I do when I encounter this problem? After the derivation

Monotonically increasing interval of function f (x) = 2x square - LNX A (0,1 / 2) B (0,4 radical 2) C (1 / 2, + infinite) d (- 1 / 2,0) and (0,1 / 2) what should I do when I encounter this problem? After the derivation

After finding the derivative, then judge which interval the derivative is greater than 0, which interval is less than 0, which interval the derivative is greater than 0, in which interval the original function increases, in which interval the derivative is less than 0, and in which interval the original function decreases. For example, this problem: F (x) = 2x ²- LNX derivative: F '(x) = 4x-1 / X let derivative f' (x) >

Find the monotone interval of function y = - LNX + 2x ^ 2

Y '= - (1 / x) + 4x (x > 0) because LNX makes sense
=(4x^2-1)/x
Let y '> 0
The increment interval is [1 / 2, + infinity)
y'

The monotone decreasing interval of function y = 2x / LNX is

The definition domain of function y is: x > 0, and X ≠ 1;
Take the derivative of the function y = 2x / LNX and get:
y'=(2lnx-2)/(lnx)^2=2/(lnx)^2*(lnx-1)
Let y '< 0 to obtain:
lnx-1＜0
I.e. LNX < 1
so
The monotonic decreasing interval of function y = 2x / LNX is:
（0,1）∪(1,e)

Curve y = x (1-ax) 2 (a > 0), and Y ′| x = 2 = 5, find the value of real number a

Y = x (1-ax) 2 = x (1-2ax + a2x2) = x-2ax2 + a2x3
Y ′ = 1-4ax + 3a2x2
Because y ′| x = 2 = 5, that is, 1-8a + 12a2 = 5 (a > 0),
The solution is a = 1

How to find the derivative of curve y = e ^ ax

You think so
Y=e^ax
y'=e^ax ▪ (AX) '(this is the derivative of the composite function)
=e^ax ▪ a
=a ▪ e^ax

Given that the straight line y = (a + 1) X-1 and the curve y2 = ax have exactly one common point, find the value of the real number a

Simultaneous equations:
y＝(a+1)x−1
y2＝ax ，
Eliminate y to obtain: ((a + 1) x-1) 2 = ax
Simplified as: (a + 1) 2x2 - (3a + 2) x + 1 = 0
① When a = - 1, it is obviously true
② When a ≠ - 1, △ = (3a + 2) 2-4 (a + 1) 2 = 0, the solution is a = 0 or − 4
five
To sum up, a = 0 or - 1 or − 4
five