Given that the value domain of function f (x) = 3x-4 is [- 10,5], find its definition domain?

Given that the value domain of function f (x) = 3x-4 is [- 10,5], find its definition domain?

Because the function field is only [- 10,5]
So - 10 ≤ 3x-4 ≤ 5
-6≤3x≤9
-2≤x≤3
Therefore, the function definition field is [- 2,3]

Given f (x + 1) = x ^ 2, find the value field of F (x) known function f (x) = 3x-4 as [- 10,5] and find its definition field

set up
t=x+1;
X = T-1, then f (T) = (t-1) ^ 2 = T ^ 2-2 * t + 1
Monotonically increasing by F (x) = 3x-4
-10

Find the definition field of function y = √ (6x-x ^ 2-5) / √ (10 + 3x-x ^ 2)

10+3x-x^2 > 0 ; 6x-x^2-5>=0
(x+2)(x-5)

Given that the domain of function y = f (3x-2) is [1 / 3,5 / 3], find the domain of y = f (3-4x)

Let x = 3x-2, X ∈ [1 / 3,5 / 3]
Then the definition field of F (x) is: X ∈ [- 1,3]
∴-1≤3-4x≤3
∴0≤x≤1.
That is, the definition field of y = f (3-4x) is [0,1]

The interval of the domain of the function y = √ (5 - |x|) / 3x + 1 is expressed as

Closed interval - 5 to 5 - 1 / 3 removed

Y = f (x) = 8 / ((x + 1) ^ 2) how to find the derivative? As the title, Sorry, I made a mistake. It should be the integral between 0 and 1

y＝8/(x+1) ²
→y＝8(x+1)^(-2)
→y'＝8 × (-2) × (x+1)^(-2-1)
→y'＝-16(x+1)^(-3).
∴y'＝-16/(x+1) ³.