Find the value range of the following function: y = x + √ (4-x) ²）

Find the value range of the following function: y = x + √ (4-x) ²）

This problem uses triangular substitution (substitution method) to make x = 2sint. Let t ∈ (- π / 2, π / 2) ‡ y = 2sint + 2|cost| = 2sint + 2cost t ∈ (- π / 2, π / 2) = 2 √ 2Sin (T + π / 4) t ∈ (- π / 2, π / 2) ∈ when t + π / 4 = π / 2, that is, t = π / 4, y obtains the maximum value of 2 √ 2. When t + π / 4 = - π / 4, that is, t = - π / 2

Find the function range y = X ²+ 2x-3 (x∈R)

The opening is upward, and the axis of symmetry is x = - 1
When x = - 1, get the minimum value 1-2-3 = - 4
So the value range is [- 4, + ∞)

Find the value range of the following functions: ① y = 3x-5 / X-2 ② y = - 2x ²- 3x-4.x∈[-2,1]

1) Y = 3x-5 / X-2 = [3 (X-2) + 1] / (X-2) = 3 + 1 / (X-2) 1 / (X-2) ≠ 0 3 + 1 / (X-2) ≠ 3. The value range is: y ≠ 32) y = - 2x ²- 3x-4=-2(x ²+ 3/2x)-4=-2(x ²+ 3/2x+9/16)+9/8-4=-2(x+3/4) ²- 23 / 8 when x = - 3 / 4, the function takes the maximum value of - 23 / 8. When x = 1, the function

Find the value range of the following functions: 1. Y = 2x-1 - √ (13-4x) 2. Y = 4 - √ (3 + 2x-x) ² Find the value range of the following functions 1.y=2x-1-√（13-4x） 2.y=4-√（3+2x-x ²） 3.y=4/（2x ²- 4x-1）

The number in the root sign should be greater than or equal to 0, so 13-4x ≥ 0, i.e. ≤ 13 / 4. Because 2x increases in (- ∞, 13 / 4] and √ (13-4x) also increases in (- ∞, 11 / 2] root sign = 3 + 2x-x ²=- (x-1) ^ 2 + 4, so 0 ≤√ (3 + 2x-x) ²) ≤ 2. So the value range of Y [2,4] y = 4 / [2 (x-1) ^ 2-5], because

High number of an inverse derivative Find the inverse derivative of x ^ 2ln (x ^ 2 + 5) DX

It is estimated that the landlord requires its points
It can be obtained by partial integral method
∫x^2*Ln(x^2+5)dx
= 2/9*x^3-10/3*x+1/3*x^3*Ln(x^2+5)-10√5/3*arctan(x/√5)+C
Where C is an arbitrary constant

A higher derivative problem! There is only one problem, G (x) = (x ^ 2) sin1 / x, X ≠ 0, x = 0, that is, a piecewise function, and f (x) is differentiable. Find the derivative of function f (x) = f (g (x)) at x = 0 My main question is that the definition of derivative in the positive solution of this problem seems to be finally reduced to f '(0) * g' (0). Wouldn't it be enough to directly find the derivative of the formula F (x)? Also, if you take the derivative of G (x), it is equal to 2xsin1 / x-cos1 / X. This formula is equal to? But if by definition g'0 = xsin1 / X is exactly 0

g(x)=(x^2)sin1/x,x≠0
By definition, G'0 = xsin1 / X is exactly 0. It indicates that there is a derivative at 0, but the derivative function is discontinuous
The formula of compound derivation requires that the derivatives in it should be continuous
(although it is not mentioned in the book, it is implied that the value can be substituted, that is, the condition of continuity of derivative function)
In this problem, the derivative of G (x) is discontinuous at 0, so we can't find the derivative by composite, but by definition
The single point derivative can only be obtained by definition. The compound derivative rule is a derivative function, not a value