Okay so I need help with basically how to find the domain, range, and how to solve if it says Find the value of f(3).
Let the f function be defined by f(x) = the square root of x + 1 for -1 is less than or equal to x less than or equal to 3.
And 4 - x for 3 is less than x less than or equal to 5
I don't know where to start. Thanks in advance.
we have:
f(x) =
{√(x+1) for -1 <= x <= 3
{4-x for 3 < x <= 5
domain:
square root is only defined for positive values. So, we need x+1 >= 0, or x >= -1. 4-x is defined for all real x.
So, domain is [-1,5]
range:
for -1 <= x <= 3, 0 <= f(x) <= 2
for 3 < x <= 5, -1 <= f(x) < 1
so, the range is [-1,2]
f(3) = √(3+1) = √4 = 2
To find the domain of a function, you need to determine all the possible values that the input variable (in this case, x) can take.
In your given function, you have two different pieces for different ranges of x. Let's break it down:
1. For -1 ≤ x ≤ 3:
- The function is defined as f(x) = √(x + 1).
- For this range, the square root function is defined for all non-negative real numbers. Since x + 1 must be non-negative, we have x + 1 ≥ 0, which gives us x ≥ -1.
- Therefore, the domain for this part is -1 ≤ x ≤ 3.
2. For 3 < x ≤ 5:
- The function is defined as f(x) = 4 - x.
- For this range, there are no additional restrictions on x. So the domain for this part is 3 < x ≤ 5.
Combining the two domains, we get the overall domain for the function: -1 ≤ x ≤ 3 or 3 < x ≤ 5.
Now, let's move on to the range. The range refers to all the possible values that the output variable (in this case, f(x)) can take.
1. For -1 ≤ x ≤ 3:
- The function is defined as f(x) = √(x + 1).
- Since x + 1 can take any non-negative value for this range, the square root function will output real numbers greater than or equal to 0.
- Therefore, the range for this part is f(x) ≥ 0.
2. For 3 < x ≤ 5:
- The function is defined as f(x) = 4 - x.
- Since x can take any value in this range, the function will output real numbers.
- Therefore, the range for this part is f(x) ∈ ℝ.
Combining the two ranges, we get the overall range for the function: f(x) ≥ 0 (for -1 ≤ x ≤ 3) and f(x) ∈ ℝ (for 3 < x ≤ 5).
Now, to solve for f(3), you need to substitute the value of x as 3 into the function.
1. For -1 ≤ x ≤ 3:
- The function is f(x) = √(x + 1).
- Substitute x = 3: f(3) = √(3 + 1).
- Evaluate: f(3) = √4 = 2.
2. For 3 < x ≤ 5:
- The function is f(x) = 4 - x.
- Substitute x = 3: f(3) = 4 - 3.
- Evaluate: f(3) = 1.
Since we have two different functions for different ranges, we need to consider each separately. Therefore, f(3) evaluates to 2 for -1 ≤ x ≤ 3 and 1 for 3 < x ≤ 5.