FX = X-2 / x + a (2-lnx) a > 0 find FX monotone interval

FX = X-2 / x + a (2-lnx) a > 0 find FX monotone interval

F '(x) = 1 + 2 / x ^ 2-A ^ 2 / x = (1-radical 2 / x) ^ 2 + (2 * radical 2-A ^ 2)
Obviously, when a ^ 20, f (x) increases monotonically
1-radical (a ^ 2-2 * radical 2)

LNX + 1 = 0, how much is x equal to?

My answer is 1 / E, and the steps are: LNX = - 1
Both sides of the equation become exponential power of e at the same time to get the - 1 power of x = E,
So the result is x = 1 / E

What is LNX = 1. X equal to?

What is limx → 1 (1 / lnx-1 / x-1) equal to

lim(x→1)　[1/lnx-1/(x-1)]
=LIM (x → 1) [x-1-lnx] / [LNX (x-1)]
＝lim(x→1)(1-1/x)/[(x-1)/x+lnx]
=LIM (x → 1) (x-1) / (x-1 + xlnx)
＝lim(x→1)1/(1+lnx+1)
=1/2
If it's useful, give it a good comment_ ∩)O~~

When x > 1, LNX + 1 / LNX is greater than or equal to 2,

Yes
Reason: when x > 1, LNX > 0
lnx+1/lnx>=2√(lnx*1/lnx)=2

LNX = 1 / 2, what is x equal to?

lnx=1/2
x=e^(1/2)
X = √ e is the root sign E

How is x ^ (1 / x) equal to e ^ (LNX / x), is there any formula

x^(1/x)=e^ln(x^(1/x))
=e^((lnx)/x)
It's a logarithmic formula

What does x → + ∞. E ^ (LNX / x) equal
If it's equal to 1, doesn't that LNX molecule matter, or does it approach 0 as long as the denominator is infinite

Because LIM (x → + ∞) LNX / x = LIM (x → + ∞) 1 / x = 0
therefore
lim(x→+∞)e^(lnx/x)=e^0=1

When x → e, find (lnx-1) / (x-e)

lim(x->e) (lnx-1)/(x-e) (0/0)= lim(x->e) (1/x)/1=1/e or expands lnx about elnx = lne +(x-e)/e + (x-e)^2/e^2+...= 1+(x-e)/e + (x-e)^2/e^2+...(lnx-1)/(x-e)= [ 1+(x-e)/e + (x-e)^2/e^2+...- 1] /(x-e)= ((x...

Why is LNX greater than or equal to negative 1 less than or equal to 0 converted to greater than or equal to 1 / E less than or equal to 1

-1