Question

range of f(x) = 3|sin x| - 4|cos x|

Answer 1

[-4, 3]

Answer 2

To find the range of the function $f(x) = 3|\sin x| - 4|\cos x|$, we can analyze the behavior of the terms $|\sin x|$ and $|\cos x|$.

1. Periodicity: The function $f(x)$ has a period of $\pi$. $f(x+\pi) = 3|\sin(x+\pi)| - 4|\cos(x+\pi)| = 3|-\sin x| - 4|-\cos x| = 3|\sin x| - 4|\cos x| = f(x)$. Therefore, we only need to analyze the function over an interval of length $\pi$, for example, $[0, \pi]$.

2. Case 1: $x \in [0, \pi/2]$ In this interval, $\sin x \ge 0$ and $\cos x \ge 0$. So, $|\sin x| = \sin x$ and $|\cos x| = \cos x$. The function becomes $f(x) = 3\sin x - 4\cos x$. Let $g(x) = 3\sin x - 4\cos x$. To find the range of $g(x)$ for $x \in [0, \pi/2]$, we can use calculus or the auxiliary angle method.

   • Using Calculus: $g'(x) = \frac{d}{dx}(3\sin x - 4\cos x) = 3\cos x + 4\sin x$. For $x \in [0, \pi/2]$, $\cos x \ge 0$ and $\sin x \ge 0$. Thus, $g'(x) \ge 0$ for all $x \in [0, \pi/2]$. This means $g(x)$ is a monotonically increasing function in this interval. The minimum value occurs at $x=0$: $g(0) = 3\sin(0) - 4\cos(0) = 3(0) - 4(1) = -4$. The maximum value occurs at $x=\pi/2$: $g(\pi/2) = 3\sin(\pi/2) - 4\cos(\pi/2) = 3(1) - 4(0) = 3$. So, for $x \in [0, \pi/2]$, the range of $f(x)$ is $[-4, 3]$.

3. Case 2: $x \in [\pi/2, \pi]$ In this interval, $\sin x \ge 0$ and $\cos x \le 0$. So, $|\sin x| = \sin x$ and $|\cos x| = -\cos x$. The function becomes $f(x) = 3\sin x - 4(-\cos x) = 3\sin x + 4\cos x$. Let $h(x) = 3\sin x + 4\cos x$. To find the range of $h(x)$ for $x \in [\pi/2, \pi]$:

   • Using Calculus: $h'(x) = \frac{d}{dx}(3\sin x + 4\cos x) = 3\cos x - 4\sin x$. For $x \in [\pi/2, \pi]$, $\cos x \le 0$ and $\sin x \ge 0$. Thus, $3\cos x \le 0$ and $-4\sin x \le 0$. So, $h'(x) \le 0$ for all $x \in [\pi/2, \pi]$. This means $h(x)$ is a monotonically decreasing function in this interval. The minimum value occurs at $x=\pi$: $h(\pi) = 3\sin(\pi) + 4\cos(\pi) = 3(0) + 4(-1) = -4$. The maximum value occurs at $x=\pi/2$: $h(\pi/2) = 3\sin(\pi/2) + 4\cos(\pi/2) = 3(1) + 4(0) = 3$. So, for $x \in [\pi/2, \pi]$, the range of $f(x)$ is $[-4, 3]$.

4. Combining the ranges: Since the function's behavior repeats every $\pi$ and the range over $[0, \pi/2]$ is $[-4, 3]$ and over $[\pi/2, \pi]$ is $[-4, 3]$, the overall range of $f(x)$ for all real $x$ is $[-4, 3]$.

Alternative Approach (Generalization using $\theta$): Let $u = |\sin x|$ and $v = |\cos x|$. We know that $u \ge 0$, $v \ge 0$ and $u^2 + v^2 = |\sin x|^2 + |\cos x|^2 = \sin^2 x + \cos^2 x = 1$. So, we are looking for the range of $3u - 4v$ where $u = \sin \theta$ and $v = \cos \theta$ for some $\theta \in [0, \pi/2]$. This is because for any $x$, the pair $(|\sin x|, |\cos x|)$ will always correspond to a pair $(\sin\theta, \cos\theta)$ for some $\theta \in [0, \pi/2]$. For example:

• If $x \in [0, \pi/2]$, take $\theta = x$.
• If $x \in [\pi/2, \pi]$, take $\theta = \pi - x$. Then $\sin\theta = \sin(\pi-x) = \sin x = |\sin x|$ and $\cos\theta = \cos(\pi-x) = -\cos x = |\cos x|$.
• If $x \in [\pi, 3\pi/2]$, take $\theta = x - \pi$. Then $\sin\theta = \sin(x-\pi) = -\sin x = |\sin x|$ and $\cos\theta = \cos(x-\pi) = -\cos x = |\cos x|$.
• If $x \in [3\pi/2, 2\pi]$, take $\theta = 2\pi - x$. Then $\sin\theta = \sin(2\pi-x) = -\sin x = |\sin x|$ and $\cos\theta = \cos(2\pi-x) = \cos x = |\cos x|$.

In all cases, the function $f(x)$ transforms into $g(\theta) = 3\sin\theta - 4\cos\theta$ for $\theta \in [0, \pi/2]$. As shown in Case 1, the range of $g(\theta)$ for $\theta \in [0, \pi/2]$ is $[-4, 3]$.

The final answer is $\boxed{[-4, 3]}$.