Find the extremum of function y = x ^ (1 / x) (to calculate the process) Those who have not studied advanced mathematics are advised not to do it. When x = 1 is y = 1, is 1 a minimum?

y=x^(1/x)
lny=lnx^(1/x)=1/xlnx
y'/y=-1/x^2lnx+1/x^2
y'=y(1/x^2-lnx/x^2)
=x^(1/x)(1/x^2-lnx/x^2)
=0
x^(1/x)>0
1/x^2-lnx/x^2=0
(1-lnx)/x^2=0
lnx=1
x=e
The extreme points are as follows
x=e
The extreme value is:
f(e)
=e^(1/e)
Why do you think it's a minimum,
Is the minimum the minimum?

Finding the extremum of function y = x square + 2x + 1

This function has only one minimum value 0, no maximum value!

Find the abscissa of the intersection of the function image of y = LNX and y = 3-x
It's a process

There is no way to find out the exact value of this problem
But we can use dichotomy to get the general range
If the intersection satisfies f (x) = LNX + x-3 = 0, the zero point of this equation is solved by dichotomy. If the zero point is in the interval (a, b), then f (a) f (b)

It is proved that the image of the function y FX is below the line L, and the function FX LNX ax 1 is known
Given that f (x) = lnx-ax + 1, a is a real number and a constant,
(1) Find the equation of the tangent l of the image with function y = f (x) at point P (1, f (1)), and prove that the image with function y = f (x) (x is not equal to 1) is below the line L;
(2) The number of zeros of function y = f (x) is discussed
The first question is y = (1-A) X

The difference between the two formulas leads to x-lnx-1. The derivation shows that the formula is more than zero, and it holds. The first question is proved
Take one as the boundary
If a = 1, there is only one solution x = 1
If A1 is derived from F (x), we can judge whether the derivative is positive or negative, which is good

Given the function f (x) = e ^ | LNX | - | X-1 / X |, then the approximate image of the function y = f (x + 1) is

Shift the image of F (x) one unit to the left

What is the X / LNX function image like
Seeking the picture

The definition field is X & gt; 0 and X ≠ 1. The image is shown in the figure

The maximum value of function y = (LNX) ^ 2 / X is

y=(ln²x)/x
Y '= (2lnx - ln & # 178; x) / X & # 178;, let y' = 0, we get 2lnx ln & # 178; X = 0, LNX = 0 or LNX = 2, x = 1 or x = E & # 178;
When 00, y is an increasing function,
So the maximum value is f (E & # 178;) = ln & # 178; E & # 178 / E & # 178; = 4 / E & # 178;

If f (x) = (1-A + LNX) / X and a is r, then the maximum of F (x) is?

f'(x)=[x/x-(1-a+lnx)]/x^2=(a-lnx)/x^2=0---> a-lnx=0---> x=e^a
Because x > e ^ A, f '

Given the function f (x) = 2F ′ (1) lnx-x, the maximum of F (x) is______ ．

Since the function f (x) = 2F ′ (1) lnx-x, then f ′ (x) = 2F ′ (1) × 1x-1 (x > 0), f ′ (1) = 2F ′ (1) - 1, so f ′ (1) = 1, f ′ (x) = 2 × 1x-1 = 2 − XX, Let f ′ (x) > 0, the solution is: x < 2, Let f ′ (x) < 0, the solution is: x > 2, then the function is an increasing function on (0,2), and a decreasing function on (2, + ∞), so the maximum of F (x) F (2) = 2ln2-2, so the answer is: 2ln2-2

The maximum of function f (x) = (a + LNX) / X (a ∈ R) is equal to?
I also understand the specific, is not ~ ~ ah ~ ~ to process!! thank you

First, find the derivative of the function f (x), and then make the derivative equal to 0. The obtained x is the maximum point, and then substitute it into the analytic expression of the function to find the f (x) at this time

Find the extremum of function y = x ^ (1 / x) (to calculate the process) Those who have not studied advanced mathematics are advised not to do it. When x = 1 is y = 1, is 1 a minimum?