The extremum of the function y = x-e Λ x is?

The extremum of the function y = x-e Λ x is?

A:
y=x-e^x
Derivation: y '= 1-e ^ x, zero point x = 0
Derivation again: y '' = - e ^ x

Function f (x) = x3-3x2 + 1 in X=______ The minimum value is obtained at

If f '(x) = 3x2-6x, Let f' (x) = 3x2-6x = 0, then X1 = 0, X2 = 2 and X ∈ (- ∞, 0), f '(x) > 0; if x ∈ (0, 2), f' (x) < 0; if x ∈ (2, + ∞), f '(x) > 0, so f (x) gets the minimum value when x = 2. So the answer is 2

It's 24:3, - 5,7, - 13

[(-5)*(-13)+7]/3=24

How to calculate the curvature of a given polar equation?
If the polar coordinates can't be calculated, it can be solved by a parametric equation, but my polar coordinates equation is y = 40-10cos2x. The Y and X in the equation represent the "meat" and the angle "Sita" in polar coordinates respectively
It's very troublesome to find the derivative in the parameter equation. If you have a master, you can use software to solve it

In polar coordinates, the formula of curvature is:
K=|ρ^2+2ρ'^2-ρρ''|/(ρ^2+ρ'^2)^(3/2)
Then put it in yourself

When selecting salt water, the density of salt water is required to be 1.1 × 10 ᦉ kg / m ᦉ 179;, and now 0.5 m ᦉ 179; salt water is prepared, and the weight is 600 kg. Does the salt water meet the requirements? If not, add salt or water? How much?

Because: 600 / 0.5 = 1200 kg / m ^ 3 > 1.1 * 10 ^ 3 kg / m ^ 3, it does not meet the requirements, the density is too high, and the water with low density should be densified
Let: add water VM ^ 3; then M add water = 1.0 * 10 ^ 3kg / m ^ 3 * VM ^ 3 = 1.0 * 10 ^ 3vkg;
There are: (600 + 1.0 * 10 ^ 3V) / (0.5 + V) = 1.1 * 10 ^ 3;
The solution is: v = 0.5 (m ^ 3), then: m water = 500kg

The front view and left view of a space geometry are equilateral triangles with side length 1, and the top view is a circle

∵ the front view and the left view of the geometry are equilateral triangles with side length of 1, the top view is a circle, ∵ the geometry should be a cone, and the radius of the bottom circle is 12, the length of the generatrix is 1, ∵ the height of the cone is 32, so the side area is 12 × π × 1 = π 2, and the volume is 13 π× (12) 2 × 32 = 312 π

If the tangent of the curve y = f (x) at P (1, Y0) is parallel to the straight line y = - x + 1, find the simple form of the function y = f (x)
Given that the function f (x) = A / x + lnx-1 a belongs to the R 1 problem, if the tangent of the curve y = f (x) at P (1, Y0) is parallel to the straight line y = - x + 1, find the monotone interval of the function y = f (x). If a > 0 and X belongs to (0,2e), f (x) is constant, find the range of the real number a

(1) f'(x)=-a/x^2+1/x
From the condition, a = - 2
f'(x)=-2/x^2+1/x=(x-2)/x^2
0

What is the difference between the dry density of soil and the density of soil
I always can't tell the difference between "wet density" and "code for highway geotechnical experiment" today

The dry density of soil is the mass of solid particles per unit volume of soil
The density and "wet density" of soil is a concept, which is the mass per unit volume of soil (including the mass of solid particles and water in soil)!
So the density of soil is higher than the dry density!

16 (x-1) square = 225 use square difference and complete square formula to solve equation

Solution
16(x-1)^2=225
(x-1)^2=225/16
x-1=±15/4
∴x=1+15/4=19/4
Or x = 1-15 / 4 = - 11 / 4

The 101st power of (- 1) + the square of 1-radical 1-radical 2
The process of seeking clarity

The 101st power of (- 1) + the square of 1-radical 1-radical 2
=-1+1-1-2
=-3