Value range of independent variable X of function y = 2x-1 / 1

Value range of independent variable X of function y = 2x-1 / 1

Not equal to 1 / 2

In function y = 1 In 2x − 1, the value range of the independent variable x is __

According to the meaning of the question: 2x-1 ≠ 0,
The solution is x ≠ 1
2．
So the answer is x ≠ 1
2．

What is the value range of the independent variable X of the function y = 2x-1 / 3?

Is y = 1 / (3x-1)?
If so, there is a saying that the denominator is not zero in the general range, so the value range should be x= 1/3.

Function y = 2x + 1, the value range of independent variable is

This is a function
The argument can take any real number, that is, R

Let the function f (x) be differentiable at x = 1, f (1) = 1, f '(1) = 3, then LIM (H tends to 0) [F2 (1 + H) - F2 (1)] / ln (1 + H) = how much

Should F2 (1 + H) written by the building group be the square of F (1 + H)? If so, it's easy to say. The solution is as follows:
F ^ 2 (1 + H) represents the square of F (1 + H)
When h → 0, ln (1 + H) is equivalent to H,
Then Lim [f ^ 2 (1 + H) - f ^ 2 (1)] / ln (1 + H)
=lim[f(1+h)+f(1)]*[f(1+h)-f(1)]/h
=lim2f(1)*f'(1)
=6

Let z = Y / (f (x ^ 2-y ^ 2)), where f is a differentiable function, verify Verification: ((1 / x) (∂ Z / ∂ x)) + ((1 / y) (∂ Z / ∂ y)) = Z / (y ^ 2)

∂ Z / ∂ x = - ((∂ f / ∂ x) * y * 2x) / f ^ 2 ∂ Z / ∂ y = 1 / F + 2Y2 * (∂ f / ∂ y) / f ^ 21 / F = Z / Y ∂ f / ∂ x = 2x * f '∂ F / ∂ y = - 2Y * f' I have listed the key steps because of formula editing