find domain:
f(x)=sqrt of 100-x^2
f(x)=sqrt of 2x^2+6x-8
Think of f(x) as being pretty much like y. The domain is what you are allowed to put in for x in order to get some f(x) or y to come out. That is called the range, but, anyway, are there any x values that you can see as being troublesome? Which x's work fine and which do not? Are you supposed to express this using a certain notation or not? Does the problem itself specify any restrictions?
To find the domain of a function, you need to determine which values of "x" are valid inputs that will not result in any undefined or forbidden results. In both of the given functions, we have square root operations, so we need to ensure that the values inside the square root are non-negative.
Function 1: f(x) = √(100 - x^2)
For the square root to be defined, the value inside it (100 - x^2) must be greater than or equal to zero. So, we solve the inequality:
100 - x^2 ≥ 0
To solve this inequality, we can factor it as (10 - x)(10 + x) ≥ 0. Now, we have two cases to consider:
Case 1: (10 - x)(10 + x) ≥ 0
When (10 - x)(10 + x) ≥ 0, it means that either both factors are positive or both factors are negative.
When 10 - x ≥ 0 and 10 + x ≥ 0:
10 - x ≥ 0 → x ≤ 10
10 + x ≥ 0 → x ≥ -10
So, the solution for this case is -10 ≤ x ≤ 10.
Case 2: (10 - x)(10 + x) < 0
When (10 - x)(10 + x) < 0, it means that one factor is positive and the other is negative.
When 10 - x < 0 and 10 + x < 0:
10 - x < 0 → x > 10
10 + x < 0 → x < -10
However, this case doesn't fulfill the condition for the square root to be defined. Therefore, we discard this case.
Thus, the domain of f(x) = √(100 - x^2) is -10 ≤ x ≤ 10.
Function 2: f(x) = √(2x^2 + 6x - 8)
To find the domain, we set the expression inside the square root to be greater than or equal to zero:
2x^2 + 6x - 8 ≥ 0
To solve this inequality, you can either factor the quadratic expression or use the quadratic formula. Let's use the quadratic formula in this case:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 6, and c = -8. Plugging these values into the formula, we have:
x = (-6 ± √(6^2 - 4(2)(-8))) / (2(2))
x = (-6 ± √(36 + 64)) / 4
x = (-6 ± √100) / 4
x = (-6 ± 10) / 4
So, x can be equal to (-6 + 10)/4 = 1 or (-6 - 10)/4 = -4.
Thus, the domain of f(x) = √(2x^2 + 6x - 8) is x = 1 and x = -4.