Find the function y = X ² Extreme value of LNX f(x)=x ² LNX derivative: f'(x)=2xlnx+x=0 I will, but I really don't understand how the value of X comes from The answer is x = 1 / √ E I just need to take the steps of "Y"

Make an appointment to go to X:
2lnx+1=0
lnx=-1/2
e^(lnx)=e^(-1/2)
So we need your answer

Find the extreme value of function y = x-lnx, X ∈ (0,2)

Y '= 1-1 / x = (x-1) / x = 0, x = 1
0

If the function f (x) = 2x2 LNX is not a monotone function in a sub interval (k-1, K + 1) in its definition domain, the value range of real number k is () A. [1，+∞） B. [1，3 2） C. [1，2） D. [3 2，2）

Because the definition field of F (x) is (0, + ∞), and f '(x) = 4x − 1
x，
From F '(x) = 0, x = 1 is obtained
2．
When x ∈ (0, 1)
2) When f '(x) ＜ 0, when x ∈ (1
2, + ∞), f '(x) ＞ 0
According to the meaning of the title,
k−1＜1
2＜k+1
k−1≥0 ，
The solution is 1 ≤ K < 3
2．
Therefore, B

If the function f (x) = 2x2 LNX is not a monotone function in a sub interval (k-1, K + 1) in its definition domain, the value range of real number k is () A. [1，+∞） B. [1，3 2） C. [1，2） D. [3 2，2）

If the function f (x) = 2x ^ 2-lnx is not a monotone function on a sub interval (k-1, K + 1) of the definition field, the value range of the real number is () A.k>3/2 B.k

First find that the reciprocal of F is 4x-1 / X
When x = 1 / 2 is obtained, it is the turning point
To make the defined subinterval not monotonic, it is necessary to include x = 1 / 2
So k-1 ＜ 1 / 2
K+1〉1/2
Obtain - 1 / 2 ＜ K ＜ 3 / 2
So choose C

After function f (x) = X-2, x under the root of the fraction ²- The definition field of 5x + 6 is?

After f (x) = X-2, x under the root of parts ²- 5x + 6 makes sense
Then X-2 ≠ 0, and X ²- 5x+6≥0
Then x ≠ 2, (X-2) (x-3) ≥ 0
Then x ≥ 3 or x < 2
Therefore, the definition domain is (- ∞, 2) ∪ [3, + ∞)

Find the function y = X ² Extreme value of LNX f(x)=x ² LNX derivative: f'(x)=2xlnx+x=0 I will, but I really don't understand how the value of X comes from The answer is x = 1 / √ E I just need to take the steps of "Y"