Let z = y ^ X be the function of two variables to get the second order partial derivative

Let z = y ^ X be the function of two variables to get the second order partial derivative

F (x) = e ^ - KX + (x ^ 2 + X-1 / k) how to derive this function is mainly e ^ - KX
Finding monotone interval of function
Is the maximum value of the real number function equal to 3E ^ - 2? Give reasons

If G (x) = e ^ H (x), then G '(x) = H' (x) e ^ H (x)
So f (x) = e ^ - KX + (x ^ 2 + X-1 / k)
f '(x) = [d(-kx)/dx]*e^(-kx) + 2x + 1
= -ke^(-kx) + 2x + 1

Derivation f (x) = kx-x / k-2lnx
Why is the middle one a plus sign

Derivative = k-1 / K-2 / X
In the middle is the minus sign

There are two points m and N on the line AB, point m divides AB into two parts am: MB = 2:3, point n divides AB into two parts an: NB = 4:3, and Mn = 3?
Don't count
I didn't learn the linear equation of two variables

Let am = 2x, then MB = 3x, ab = am + MB = 5x
AN=AM+MN=2x+3
4 / 7 of an = AB is obtained from an: NB = 4:3
That is, 2x + 3 = 5x * (4 / 7)
x=3.5
AB=5x=5*3.5=17.5

It is known that there are two points m and N on the line ab. point m divides AB into two parts 2:3 and point n divides AB into two parts 4:1. If Mn = 8cm, the length of AM and Nb can be obtained

AM:MB=2:3
AM=2/5AB
AN:NB=4:1
NB=1/5AB
MN=AB-AM-NB
=AB-2/5AB-1/5AB
=2/5AB
AB=5/2MN
=5/2*8
=20
Then am = 2 / 5ab = 2 / 5 * 20 = 8
NB=1/5AB=1/5*20=4

k

Only when the opening of the image is downward can the quadratic function have the maximum value, and the coefficient of the quadratic term is negative
-(k²+k-2)0
k1
So K

If - 4 < x < 1, then the maximum value of y = (X & # 178; - 2x + 2) / (2x-2) is_____

Y = (x ^ 2-2x + 2) / (2x-2) = [(x-1) ^ 2 + 1] / 2 (x-1) = [(x-1) + 1 / (x-1)] / 2
Because - 4

Find the maximum value of y = - X & # 178; - 1 / 2x + 2

y=-x²-(1/2)x+2
=-(x²+ x/2)+2
=-(x² +x/2 +1/16) +33/16
=-(x +1/4)² +33/16
When x = - 1 / 4, y has the maximum value ymax = 33 / 16

The maximum value of F (x) = x2 + 2x + 1, X ∈ [- 2, 2] is______ ．

∵ f (x) = x2 + 2x + 1, ∵ opening up, axis of symmetry x = - 1, ∵ opening up, the farther the quadratic function is from the axis of symmetry, the greater the value of function ∵ f (x) on [- 2, 2] is the maximum value of F (2) = 9, so the answer is & nbsp; 9

Find the maximum value of (x-1) / 2x & #
The range of X is zero to positive infinity

y=(x-1)/2x²
2yx²-x+1=0
When y ≠ 0, △ = (- 1) &# 178; - 8y ≥ 0, y