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.