Question

The range of the function $f(x)=\sqrt{x^{2}+4x+4}$ is:

• A. [0, $\infty$)
• B. [1, $\infty$)
• C. [3, $\infty$)
• D. [2, $\infty$)
• E. [4, $\infty$)

Hint:

The output of a principal square root function is always non-negative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The range of a function is the set of all possible output values (y-values) that the function can produce. We need to find the set of all possible values of \( f(x) \).
Step 2: Key Formula or Approach:
The key to solving this problem is to simplify the expression inside the square root. The expression \( x^2 + 4x + 4 \) is a perfect square.
We will use the algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \) and the property \( \sqrt{y^2} = |y| \).
Step 3: Detailed Explanation:
First, simplify the function \( f(x) \).
The expression inside the square root is \( x^2 + 4x + 4 \). We can recognize this as the expansion of \( (x+2)^2 \).
\[ f(x) = \sqrt{(x+2)^2} \] The square root of a squared expression is its absolute value.
\[ f(x) = |x+2| \] Now, we need to find the range of the function \( f(x) = |x+2| \). The absolute value of any real number is always non-negative.
The minimum value of \( |y| \) is 0, which occurs when \( y = 0 \).
In our case, the minimum value of \( |x+2| \) is 0, which occurs when \( x+2 = 0 \), i.e., at \( x = -2 \).
For any other value of x, \( x+2 \) will be non-zero, and \( |x+2| \) will be a positive number.
Thus, the function \( f(x) \) can take any value from 0 to positive infinity.
Step 4: Final Answer:
The set of all possible output values is \( [0, \infty) \). Therefore, the range of the function is \( [0, \infty) \).