The maximum value of the function y = 2x ^ - 2 in the interval [1,3] is_____

The maximum value of the function y = 2x ^ - 2 in the interval [1,3] is_____

A:
y=2x^(-2)=2/x²
one

The known function f (x) = alnx + 1 / 2x ^ 2 + X
Find the monotone interval of (1) f (x) (2) function g (x) = 2 / 3x ^ 3 + X-1 / 6 (x > 0). When a = 1, the image of F (x) is not above g (x)

f'(x)=a/x+x+1=(x²+x+a)/x (x>0)
Let H (x) = x & # 178; + X + A, its axis of symmetry is x = - 1 / 2, H (0) = a
When a ≥ 0, H (x) > 0, f '(x) = H (x) / x > 0
The increasing interval (0, + ∞) of F (x)
When A0, K (x) increases
When x = 1, K (x) is the minimum
k(x)min=k(1)=2/3-1/6-0-1/2=0
∴k(x)≥0
That is, f (x) ≤ g (x) is constant
The image of F (x) is not above g (x)

The function f (x) = x2 + 2x + alnx (a ∈ R) is known. ① when a = - 4, find the minimum value of F (x); ② if the function f (x) is a monotone function in the interval (0,1), find the value range of real number a; ③ when t ≥ 1, the inequality f (2t-1) ≥ 2F (T) - 3 holds, and find the value range of real number a

① ∵ f (x) = x2 + 2x-4lnx (x ＞ 0) ∵ f '(x) = 2x + 2 − 4x = 2 (x + 2) (x − 1) x (2 points) when x ＞ 1, f' (x) ＞ 0; when 0 ＜ x ＜ 1, f '(x) ＜ 0 ∵ f (x) decreases on (0,1) and increases monotonically on (1, + ∞) ∵ f (x) min = f (1) = 3 (4 points) ② f' (x) = 2

Given the function f (x) = 2x + 2 / x + alnx, a ∈ R
(1) If the function f (x) is monotonically increasing on [1, positive infinity], find the value range of real number a
(2) Note the function g (x) = x & # 178; [f '(x) + 2x-2]. If the minimum value of G (x) is - 6, find the minimum value of function f (x)

The definition field of F (x) is x > 0f '(x) = 2-2 / X & # - 178; + A / x = (2x & # - 178; + AX-2) / X & # - 178; from the topic meaning: F' (x) ≥ 0 for X ∈ [1, positive infinity) constant into immediate 2x & # - 178; + AX-2 ≥ 0 for X ∈ [1, positive infinity) constant separable variable: ax ≥ - 2x & # - 178; + 2 x > 0 can be divided by Xa ≥ - 2x + 2 / x, let g (x) = - 2X + 2

How to derive derivatives with absolute values
【1】y=|sin^2 x|+cos|-x^3|
[2]y=|x^3a|-|asin^[ax]x|
[3]y=|x^3+2x+18|/|x^5-3ax+|-bx| |
I just can't solve these problems,
There is arc in every question////
What's wrong

You can try to use the piecewise derivative for this kind of derivative. It may be troublesome, and it may not be all applicable

The function y =! X! (x-1) Note: the absolute value can't be typed out. Replace with! Is there a derivative at X. = 0? If so, find out its derivative
The function y =! X! (x-1) Note: the absolute value can't be typed. Do you have a derivative at X. = 0? If so, what's the meaning of solving the derivative? Analytically, if the function is differentiable, the left and right limits at the inflection point of the image are the same, that is, the left derivative equals the right derivative, What's the meaning? What's the meaning of function derivation? How to do this kind of problem? Please master to solve it one by one. Please be more detailed
What does it mean that a function can be derived here? Thank you in detail. What I want is to understand this problem, not the steps of this problem

When y = - x (x-1) = - x square + X in the interval of -- ∞ - 0, that is, y derivative = - 2x + 1: x = 0, y = 0, y derivative = 0: in the interval of 0 -- ∞, y = x (x-1) = x square - x, that is, y derivative = 2x-1. Therefore, when x = 0, Y values are equal to 0, that is, the left and right limits are the same, but the left and right derivatives are 1 and - 1, so they are not differentiable

How to use derivative to make function graph y = x-2arctanx

y'=1-2/(1+x²)=(x²-1)/(x²+1)
x

The derivation of implicit function by MATLAB,
1. Let y = f (x) be a function determined by the equation sin ((x) + y ^ (2)) = x ^ (2) y, and find y '
2. Y = f (x) is a function determined by the equation E ^ (x + y) + YLN (x + 1) = cos2x
3. Let the function y = f (x) be determined by the equation y = 1-e ^ (y), and find dy / DX
4. Let y = f (x) be determined by the equation x (1 + y ^ (2)) - ln (x ^ (2) + y) = 0, and find y '(0)
5. Let y = f (x) be a function determined by the equation cos ^ (2) (x ^ (2) + y) = x, and find f '(x)

F is an implicit function, and the derivative formula F '(x) = - FX / FY is used
FX is the derivative of F to x, FY is the derivative of F to y
for example
syms x y
f=y-x^2
fx=diff(f,x)
fy=diff(f,y)
dydx=-fx/fy
The running results are as follows
f =
y - x^2
fx =
(-2)*x
fy =
one
dydx =
2*x

It is known that f (x) is an even function on R. when x ≥ 0, f (x) = 2 ^ X-2 √ x, and a is the positive zero of the function g (x) = ln (x + 1) - 2 / X
Then the size relation of F (- 2), f (a), f (1.5) is

From the image, when x > 1, 2 ^ x > 2 √ X and the distance is larger and larger, so f (x) > 0 and increasing
Ln (x + 1) and 2 / X have only one intersection in the first quadrant, e is about 2.78, ln (2 + 1) > 1 = 2 / 2, ln (1.5 + 1)

Is there any relationship between higher derivative and Taylor formula? If so, what is the relationship?

Taylor formula: F (x) = f (x0) + F '(x0) (x-x0) / 2! + F' '(x0) (x-x0) ^ 2 / 3! +
That is to say, Taylor formula has higher derivative