The range of function $f(x) = 4x^2 + 12x + 7, x \in \mathbb{R}$ is

The function $f(x) = 4x^2 + 12x + 7$ is a quadratic function where the coefficient of $x^2$ is positive.

Since the coefficient of $x^2$ is $+4$ (positive), the parabola opens upward, meaning the function has a minimum value.

To find the minimum value, complete the square:

$f(x) = 4x^2 + 12x + 7$

$f(x) = 4(x^2 + 3x) + 7$

$f(x) = 4(x^2 + 3x + \frac{9}{4} - \frac{9}{4}) + 7$

$f(x) = 4(x^2 + 3x + \frac{9}{4}) - 4(\frac{9}{4}) + 7$

$f(x) = 4(x + \frac{3}{2})^2 - 9 + 7$

$f(x) = 4(x + \frac{3}{2})^2 - 2$

From the completed square form:

The term $(x + \frac{3}{2})^2 \geq 0$ for all real $x$.

Therefore, $4(x + \frac{3}{2})^2 \geq 0$ for all real $x$.

This gives $f(x) = 4(x + \frac{3}{2})^2 - 2 \geq -2$.

The minimum value is $-2$, occurring when $x = -\frac{3}{2}$.

Since the parabola opens upward, the function takes all values from $-2$ to infinity.

Therefore, the range of $f(x)$ is $[-2, \infty)$.