The minimum value of the function $f(x) = 3\sin x - 4\cos x, x \in [-4\pi, 4\pi]$ is equal to

Given: $f(x) = 3\sin x - 4\cos x$ where $x \in [-4\pi, 4\pi]$

For any function of the form $a\sin x + b\cos x$, the maximum and minimum values can be found using:

Maximum value $= \sqrt{a^2 + b^2}$

Minimum value $= -\sqrt{a^2 + b^2}$

Identifying the coefficients:

$a = 3$ (coefficient of $\sin x$)

$b = -4$ (coefficient of $\cos x$)

Calculate $\sqrt{a^2 + b^2}$:

$\sqrt{a^2 + b^2} = \sqrt{3^2 + (-4)^2}$

$= \sqrt{9 + 16}$

$= \sqrt{25}$

$= 5$

The function $f(x) = 3\sin x - 4\cos x$ can be rewritten as $R\sin(x + \phi)$ where $R = \sqrt{a^2 + b^2} = 5$.

Since $\sin(x + \phi)$ ranges from $-1$ to $+1$, the function $f(x)$ ranges from $-5$ to $+5$.

The interval $[-4\pi, 4\pi]$ spans $8\pi$ units, which covers 4 complete cycles of the trigonometric functions (one cycle $= 2\pi$). This is sufficient for the function to achieve both its maximum and minimum values.

Maximum value of $f(x) = 5$

Minimum value of $f(x) = -5$

Therefore, the minimum value of the function is $-5$.