Finding the Least Value of \(f(x) = \cos 2x + x\) on \([-\pi/2, \pi/2]\)

We are asked to find the minimum value of the function \(f(x) = \cos 2x + x\) over the closed interval \([-\pi/2, \pi/2]\). To find the minimum value of a continuous function on a closed interval, we need to evaluate the function at its critical points within the interval and at the endpoints of the interval. The smallest of these values will be the absolute minimum.

Step 1: Find the Derivative of the Function

First, we compute the derivative of \(f(x)\) with respect to \(x\). The derivative tells us the slope of the function at any given point.

Given \(f(x) = \cos 2x + x\), the derivative \(f'(x)\) is:

Using the chain rule for \(\cos 2x\), we get \(\frac{d}{dx}(\cos 2x) = -(\sin 2x) \cdot 2 = -2 \sin 2x\). The derivative of \(x\) is \(1\).

So, the derivative is:

Step 2: Find the Critical Points

Critical points occur where the derivative is either zero or undefined. Since \(f'(x) = -2 \sin 2x + 1\) is defined for all real numbers \(x\), we only need to find where \(f'(x) = 0\).

Set \(f'(x) = 0\):

Rearranging the equation:

Now we need to solve this equation for \(x\) within the given interval \([-\pi/2, \pi/2]\). This means \(2x\) must be in the interval \([-\pi, \pi]\).

The values of an angle \(\theta\) for which \(\sin \theta = 1/2\) are \(\theta = \frac{\pi}{6}\) and \(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).

So, we have two possibilities for \(2x\) within the interval \([-\pi, \pi]\):

1. \(2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}\)
2. \(2x = \frac{5\pi}{6} \implies x = \frac{5\pi}{12}\)

Both \(x = \frac{\pi}{12}\) and \(x = \frac{5\pi}{12}\) lie within the interval \([-\pi/2, \pi/2]\). These are our critical points.

Step 3: Evaluate the Function at Critical Points and Endpoints

Now we evaluate the function \(f(x) = \cos 2x + x\) at the critical points (\(x = \pi/12, x = 5\pi/12\)) and the endpoints of the interval (\(x = -\pi/2, x = \pi/2\)).

• At the left endpoint, \(x = -\pi/2\):
  \(f(-\pi/2) = \cos(2 \cdot (-\pi/2)) + (-\pi/2) = \cos(-\pi) - \frac{\pi}{2} = -1 - \frac{\pi}{2}\)
• At the critical point, \(x = \pi/12\):
  \(f(\pi/12) = \cos(2 \cdot \pi/12) + \pi/12 = \cos(\pi/6) + \frac{\pi}{12} = \frac{\sqrt{3}}{2} + \frac{\pi}{12}\)
• At the critical point, \(x = 5\pi/12\):
  \(f(5\pi/12) = \cos(2 \cdot 5\pi/12) + 5\pi/12 = \cos(5\pi/6) + \frac{5\pi}{12} = -\frac{\sqrt{3}}{2} + \frac{5\pi}{12}\)
• At the right endpoint, \(x = \pi/2\):
  \(f(\pi/2) = \cos(2 \cdot \pi/2) + \pi/2 = \cos(\pi) + \frac{\pi}{2} = -1 + \frac{\pi}{2}\)

Step 4: Compare the Values to Find the Least Value

Let's compare the values we calculated:

By comparing the approximate values, we can see that the smallest value is approximately \(-2.5708\), which corresponds to \(f(-\pi/2) = -1 - \frac{\pi}{2}\).

Therefore, the least value of \(f(x)\) on the interval \([-\pi/2, \pi/2]\) is \(-1 - \frac{\pi}{2}\).