Q.Given $x > 0$, the values of $f(x) = -3\cos\sqrt{3 + x + x^2}$ lie in the interval ______.

The key idea is to determine the range of a composite function by finding the range of its inner components sequentially.

1. For $x > 0$, consider the inner quadratic expression $g(x) = x^2 + x + 3$. The vertex of this parabola is at $x = -1/(2 \cdot 1) = -1/2$. Since $x > 0$, the function $g(x)$ is strictly increasing on its domain. As $x \to 0^+$, $g(x) \to 3$. As $x \to \infty$, $g(x) \to \infty$. Thus, the range of $x^2 + x + 3$ for $x > 0$ is $(3, \infty)$.
2. Next, consider $\sqrt{x^2 + x + 3}$. Since $x^2 + x + 3 \in (3, \infty)$, the range of $\sqrt{x^2 + x + 3}$ is $(\sqrt{3}, \infty)$. …

To determine the range of a composite function like $f(x) = -3\cos\sqrt{3 + x + x^2}$, we work from the inside out. This means we first find the range of the innermost expression, then the range of the function applied to that result, and so on, until we reach the outermost function. This systematic approach ensures we correctly account for how each transformation affects the possible output values.

Here, the structure is:

1. A quadratic expression: $q(x) = 3 + x + x^2$.
2. A square root function: $\sqrt{q(x)}$.
3. A cosine function: $\cos(\sqrt{q(x)})$.
4. A scalar multiplication: $-3 \times \cos(\sqrt{q(x)})$.

Let's break it down step by step.

1. Determine the range of the quadratic expression $q(x) = 3 + x + x^2$ for $x > 0$.

   This is a parabola opening upwards, as the coefficient of $x^2$ is $1$ (positive).

   The vertex of the parabola is at $x = -\frac{b}{2a} = -\frac{1}{2(1)} = -\frac{1}{2}$.

   Since the given domain is $x > 0$, the vertex $x = -1/2$ is not included in our domain. For $x > 0$, the function $q(x)$ is strictly increasing because the vertex is to the left of $x=0$.

   As $x$ approaches $0$ from the positive side ($x \to 0^+$), $q(x)$ approaches $3 + 0 + 0 = 3$.

   As $x$ increases without bound ($x \to \infty$), $q(x)$ also increases without bound ($q(x) \to \infty$).

   Therefore, for $x > 0$, the range of $q(x) = 3 + x + x^2$ is $(3, \infty)$.

2. Determine the range of $\sqrt{q(x)} = \sqrt{3 + x + x^2}$.

   Since $q(x) \in (3, \infty)$, we take the square root of this interval.

   The square root function is strictly increasing for non-negative inputs.

   So, if $q(x) \in (3, \infty)$, then $\sqrt{q(x)} \in (\sqrt{3}, \infty)$.

   Let $u = \sqrt{3 + x + x^2}$. So, $u \in (\sqrt{3}, \infty)$.

   Note that $\sqrt{3} \approx 1.732$ radians.

3. Determine the range of $\cos(u)$ where $u \in (\sqrt{3}, \infty)$.

   The cosine function, $\cos(\theta)$, has a range of $[-1, 1]$ for all real $\theta$.

   Our argument $u$ belongs to the interval $(\sqrt{3}, \infty)$. This interval is quite large, extending infinitely.

   Since $\sqrt{3} \approx 1.732$ radians, which is greater than $\pi/2 \approx 1.57$ radians, the interval $(\sqrt{3}, \infty)$ covers multiple full cycles of the cosine function. …