Question

F(x)=1/√9-x2+2/x-1+√x+3 область определения​

Answer 1

Ответ:

f(x)=1/√9-x2+2/x-1+√x+3 равно f(x)=2+sqrt9^(-1)+sqrtx+2*x^(-1)-x2 OR f(x)=2+sqrt9^(-1)+sqrtx+2*x^(-1)-x2

Answer 2

To determine the domain of the function F(x), we need to identify the values of x for which the function is defined and doesn't result in any division by zero or taking the square root of a negative number.

Let's break down the function F(x) step by step:

F(x) = 1/√(9 - x^2) + 2/(x - 1) + √(x + 3)

1. The square root (√) is defined for non-negative real numbers, so √(9 - x^2) is defined when 9 - x^2 ≥ 0. This leads to x^2 ≤ 9, which means -3 ≤ x ≤ 3.

2. The fraction 1/(x - 1) is defined as long as the denominator is not equal to 0. So, x - 1 ≠ 0, which gives us x ≠ 1.

3. The square root √(x + 3) is defined for x + 3 ≥ 0, which implies x ≥ -3.

Putting it all together, the domain of the function F(x) is the intersection of the valid intervals:

-3 ≤ x ≤ 3 (from the square root √(9 - x^2)) x ≠ 1 (from the fraction 1/(x - 1)) x ≥ -3 (from the square root √(x + 3))

So, the domain of F(x) is -3 ≤ x < 1 and 1 < x ≤ 3. In interval notation, the domain can be represented as:

Domain: (-3, 1) ∪ (1, 3]