 45. The image of the function 𝑓: 𝑅 → 𝑅 defined by 𝑓 𝑥 = 5𝑥
2 + 1 is

To find the image of the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 5x^2 + 1 \), we need to determine the set of all possible output values of the function.

1. **Understanding the function:**  
   The given function is \( f(x) = 5x^2 + 1 \). Notice that \( x^2 \) is always non-negative for any real number \( x \), i.e., \( x^2 \geq 0 \). This implies:
   \[
   5x^2 \geq 0
   \]
   Adding 1 to both sides:
   \[
   5x^2 + 1 \geq 1
   \]

2. **Range of the function (Image):**  
   Since \( f(x) = 5x^2 + 1 \), the minimum value of \( f(x) \) occurs when \( x = 0 \). At \( x = 0 \):
   \[
   f(0) = 5(0)^2 + 1 = 1
   \]
   As \( x \) moves away from 0, \( f(x) \) increases. Therefore, the image of the function is the set of all real numbers greater than or equal to 1.

Thus, the image of the function \( f(x) = 5x^2 + 1 \) is:
\[
\boxed{[1, \infty)}
\]

Would you like a more detailed explanation or any further clarifications?

Here are 5 related questions you might explore:
1. What is the image of the function \( f(x) = x^2 + 2x + 3 \)?
2. How does the image change if the function is \( f(x) = -5x^2 + 1 \)?
3. What is the pre-image of the value \( y = 6 \) for this function?
4. How would you determine the image of \( f(x) = ax^2 + bx + c \) for arbitrary constants \( a \), \( b \), and \( c \)?
5. What are the domain and image of the function \( f(x) = \sqrt{5x^2 + 1} \)?

**Tip:** To find the image of a quadratic function, focus on identifying the minimum or maximum value and how the function behaves as \( x \) increases or decreases.

The image of the function 𝑓: 𝑅 → 𝑅 defined by 𝑓 𝑥 = 5𝑥^2 + 1 is

Discover how to find the image (range) of the quadratic function f(x) = 5x^2 + 1. Learn about quadratic functions, key formulas, and the range theorem. This solution explains how the function behaves and identifies its minimum value. Suitable for students in Grades 9-11.

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