Question

Let \(f:R→R\) be a function defined by \(f(x)=x^2+9\).The range of \(f \) is 

• A. \(R\)
• B. \((-∞,-9]∪[9,∞)\)
• C. \([9,∞)\)
• D. \([3,∞) \)
• E. \([3,∞)\text{U}(-∞,-3]\)

Answer 1

The Correct Option is C

We are given the function \( f(x) = x^2 + 9 \). To find its range:

Step 1: Analyze the function

The function \( f(x) = x^2 + 9 \) is a quadratic function opening upwards with its vertex at \( (0, 9) \).

Step 2: Determine the minimum value

The minimum value of \( f(x) \) occurs at \( x = 0 \):

\[ f(0) = 0^2 + 9 = 9 \]

Step 3: Determine the range

Since \( x^2 \geq 0 \) for all \( x \in \mathbb{R} \), \( f(x) \geq 9 \). Thus, the range of \( f \) is all real numbers \( y \) such that \( y \geq 9 \).

The range of \( f \) is \( [9, \infty) \), which corresponds to option (C).

Answer 2

Step 1: Understand the problem and given function.

We are tasked with finding the range of the function \( f(x) = x^2 + 9 \), where \( f: \mathbb{R} \to \mathbb{R} \).

Step 2: Analyze the function.

The function \( f(x) = x^2 + 9 \) is a quadratic function. The term \( x^2 \) is always non-negative for all real \( x \), i.e., \( x^2 \geq 0 \). Adding 9 to \( x^2 \) shifts the graph of \( x^2 \) vertically upward by 9 units.

Thus, the minimum value of \( f(x) \) occurs when \( x^2 = 0 \), which gives:

\[ f(x) = 0 + 9 = 9. \]

For all other values of \( x \), \( x^2 > 0 \), so \( f(x) > 9 \).

Step 3: Determine the range.

Since \( x^2 \geq 0 \), the smallest value of \( f(x) \) is 9, and \( f(x) \) can take any value greater than or equal to 9 as \( x^2 \) increases. Therefore, the range of \( f(x) \) is:

\[ [9, \infty). \]

Final Answer:

\( [9, \infty) \)