what is the largest domain of the following functions
f(x)=3x-5 f(x)=square root of (x+4)
f(x)= 2 over x-4 f(x)= sqaure root
of (x2-9)
f(x)= 2 over x(x+7)
Hopefully you've solved these by now. I just noticed the question and see there's no reply.
The question is
"what is the largest domain of the following functions "
(1)f(x) = 3x-5
(2)f(x) =sqrt(x+4)
(3)f(x)= 2/(x-4)
(3)f(x)= sqrt(x^2-9)
(4)f(x)= 2/(x(x+7))
For (1) I think you can see that there are no restrictions on x for the domain and the range is all reals.
For (2) you should see that the domain must be non-negative). Thus x => -4 is the domain; the range is 0 and the positive reals.
The 3rd problem requires you to see where x^2-9=>0. This is the same as |x|=>3 or x=<-3 or x=>3. The range is again non-neg. reals.
For (4) we cannot have zero in the denominator, thus the domain for x is all reals except 0 and -7. Calculating the range will probably be a challenge here. I'll let you prove that f(x) can never = 0. There is also a small interval of negative values the range doesn't take on.
...and just now I see that the question in only seeking the domain values...ok. Start a new thread if you need further assistance.
To find the largest domain for each of the given functions, we need to identify any values of x that would result in undefined or imaginary outputs. Let's go through each function one by one.
1. f(x) = 3x - 5
The function is a linear function and can be evaluated for any real value of x. Therefore, the domain of this function is all real numbers.
2. f(x) = √(x + 4)
To find the domain, we need to determine the values of x that would result in a non-real output. Since we cannot take the square root of a negative number and the expression inside the square root is x + 4, the domain will be all values of x that make x + 4 greater than or equal to 0. So, x + 4 ≥ 0 implies x ≥ -4. Therefore, the domain of this function is x ≥ -4.
3. f(x) = 2/(x - 4)
The denominator of the fraction cannot be zero as it would result in division by zero, making the expression undefined. So, we need to find the value of x that makes the denominator equal to zero. Solving x - 4 = 0 gives x = 4. Therefore, the domain of this function is all real numbers except x = 4.
4. f(x) = √(x^2 - 9)
Similar to the second function, we need to determine the values of x that would result in a non-real output. Since we cannot take the square root of a negative number and the expression inside the square root is x^2 - 9, we need to find the values of x that make x^2 - 9 greater than or equal to 0. Factoring x^2 - 9 gives (x + 3)(x - 3) ≥ 0. The graph of this expression shows that it is non-negative or zero when x ≤ -3 or x ≥ 3. Therefore, the domain of this function is x ≤ -3 or x ≥ 3.
5. f(x) = 2/(x(x + 7))
Similar to the third function, we need to find the values of x that make the denominator equal to zero. Solving x(x + 7) = 0 gives x = 0 or x = -7. Therefore, the domain of this function is all real numbers except x = 0 and x = -7.
In summary:
1. f(x) = 3x - 5 has a domain of all real numbers.
2. f(x) = √(x + 4) has a domain of x ≥ -4.
3. f(x) = 2/(x - 4) has a domain of all real numbers except x = 4.
4. f(x) = √(x^2 - 9) has a domain of x ≤ -3 or x ≥ 3.
5. f(x) = 2/(x(x + 7)) has a domain of all real numbers except x = 0 and x = -7.