Finding the extremum of the function y = x? LNX

Finding the extremum of the function y = x? LNX

The definition domain of the function y = x? LNX is x > 0
The derivative of F (x) = x? LNX is obtained
f'(x)=2xlnx+x>0
From 2lnx + 1 > 0, x > 1 / √ e is obtained
So f (x) decreases in the interval (0,1 / √ E) and increases in the interval (1 / √ e, + ∞)
So when x = 1 / √ e, the minimum value of F (x) is - 1 / (2e)

Finding the extremum of function y = LNX / X

y=lnx/x
therefore
y'=[(1/x)x-lnx]/x^2
Let y '= 0
So 1-lnx = 0
So x = E
So the extremum is f (E) = 1 / E

If the function y = xlnx-ax2 has two extreme points, then the range of real number a is______ .

According to the meaning of the title, y '= LNX + 1-2ax, if f' (x) = lnx-2ax + 1 = 0, LNX = 2ax-1, y = xlnx-ax2 has two extreme points, which is equivalent to that f '(x) = lnx-2ax + 1 has two zeros, which is equivalent to that the image of function y = LNX and y = 2ax-1 has two intersections

Let f (x) = 1 / 2aX 2 + (1-A) x-lnx, where a > - 1, if f (x) has two extreme points 1. Find the value range of real number a 2, when - 1

(1) The domain of F (x) is x > 0
The derivative of F (x) = ax + 1 - A - 1 / X
ax+1- a - 1/x=0
ax^2+(1- a)x - 1=0
(x- 1)(ax+1)=0
x1=1,x2=- 1/a
- 1/a> 0
So a

Given the function f (x) = x ^ 2 + x-lnx (x > 0), find the extremum of function f (x)

Derivation:
f'(x)=2x+1-1/x
When f '(x) = 0, x = 1 / 2
And X0
So the minimum value of F (x) is f (1 / 2) = 3 / 4 + LN2, and there is no maximum value
The answer is looking forward to your approval

Let the function f (x) = 2 / x + LNX, then find the extremum of F (x)

f'(x)=-2/x²+1/x=(x-2)/x²
The definition is x > 0
So 0

Find the function f (x) = 2 / x + LNX's extremum

The derivative becomes - 2 / x 2 + 1 / x = x - 2 / x 2, so the extreme point is f (2)

Let f (x) = x / a-lnx find the extremum of F (x)

f'(x)=1/a-1/x

Let f (x) = ax-1-lnx (a ∈ R). (I) discuss the number of extremum points of function f (x) in the domain of definition; (II) if the function f (x) is of function f (x), the number of extremum points of function f (x) is discussed (II) if the function f (x) obtains the extremum at x = 1 and holds for ∀ x ∈ (0, + ∞), f (x) ≥ bx-2, find the value range of real number B; (III) when E-1 < y < x, try to prove that e ^ (X-Y) > {ln (x + 1)} / {ln (y + 1)}

The function definition domain is x > 0, the derivation of function f (x) shows that f '(x) = A-1 / x, and the extremum point is f' (x) = 0 = A-1 / x, that is, x = 1 / a (1). Discussion: when a ≤ 0, f '(x) 0, f (x) obtains the extreme value at x = 1 / A, that is, the number of extreme points is 1. (2) if the function obtains the extremum at x = 1, then a = 1, f (x) = x-1-lnxf (x) ≥ bx-2 is tenable

The known function f (x) = 1 / 2 (x-1) ^ 2 + LNX ax + A. (1) if a = 3 / 2, find the extremum of function f (x) (2) if any x ∈ (1,3), f (x) > 0 holds. The second question of high school mathematics is solved in detail

Afds564, (1) f ′ (x) = x + 1 / X-5 / 2 = 2x2-5x + 2 / 2x, f '(x) = 0, X1 = 1 / 2, or x2 = 2. According to the analysis of function properties, the function f (x) obtains the maximum value at x = 1 / 2, f (1 / 2) = 7 / 8-ln2, the function f (x) obtains the minimum value at x = 2, f (2) = ln2-1 (2): F' (x) = x + 1 / X - (1 + a), X ∈