The slope of the tangent to the curve $y = x^2 - 4\cos x$, $x \in [0, \pi]$ is maximum when $x$ is equal to

Let the equation of the curve be $y = f(x)$.

$y = x^2 - 4\cos x$

The slope of the tangent to the curve at any point $x$ is given by the first derivative, $\frac{dy}{dx}$. Let the slope be denoted by $m$.

$m = \frac{dy}{dx}$

$m = \frac{d}{dx}(x^2 - 4\cos x)$

$m = 2x - 4(-\sin x)$

$m = 2x + 4\sin x$

To maximize this function $m$ on the interval $[0, \pi]$, differentiate $m$ with respect to $x$ and equate it to zero.

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$\frac{dm}{dx} = \frac{d}{dx}(2x + 4\sin x)$

$\frac{dm}{dx} = 2 + 4\cos x$

Setting $\frac{dm}{dx} = 0$ for critical points:

$2 + 4\cos x = 0$

$4\cos x = -2$

$\cos x = -\frac{1}{2}$

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Finding $x$ in the interval $[0, \pi]$ such that $\cos x = -\frac{1}{2}$:

Since $\cos x$ is negative, $x$ must be in the second quadrant.

Given that $\cos(\frac{\pi}{3}) = \frac{1}{2}$:

$x = \pi - \frac{\pi}{3}$

$x = \frac{2\pi}{3}$

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To confirm that this value of $x$ gives a maximum slope, find the second derivative of $m$:

$\frac{d^2m}{dx^2} = \frac{d}{dx}(2 + 4\cos x)$

$\frac{d^2m}{dx^2} = -4\sin x$

Evaluating at $x = \frac{2\pi}{3}$:

$\frac{d^2m}{dx^2}\bigg|_{x=\frac{2\pi}{3}} = -4\sin\left(\frac{2\pi}{3}\right)$

$\frac{d^2m}{dx^2}\bigg|_{x=\frac{2\pi}{3}} = -4\left(\frac{\sqrt{3}}{2}\right)$

$\frac{d^2m}{dx^2}\bigg|_{x=\frac{2\pi}{3}} = -2\sqrt{3}$

Since the second derivative is negative ($< 0$), the slope $m$ is at a local maximum at $x = \frac{2\pi}{3}$.

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Checking the endpoints of the interval $[0, \pi]$ to ensure it is the global maximum:

At $x = 0$:

$m = 2(0) + 4\sin(0)$

$m = 0$

At $x = \pi$:

$m = 2\pi + 4\sin(\pi)$

$m = 2\pi \approx 6.28$

At $x = \frac{2\pi}{3}$:

$m = 2\left(\frac{2\pi}{3}\right) + 4\left(\frac{\sqrt{3}}{2}\right)$

$m = \frac{4\pi}{3} + 2\sqrt{3} \approx 7.65$

The value at $x = \frac{2\pi}{3}$ is the highest.

Therefore, the slope of the tangent to the curve is maximum when $x = \frac{2\pi}{3}$.