The range of function $f(x) = 4x^2 + 12x+7, x∈ R$ is

$[−2,∞)$

The correct answer is Option (1) → $[−2,∞)$

Given function:

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

Rewrite in vertex form:

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

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

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

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

The minimum value occurs at $x = -\frac{3}{2}$.

Minimum value of $f(x)$ is $-2$.

Since the coefficient of $x^2$ is positive, the parabola opens upwards.

Range of $f(x)$ = $[-2, \infty)$