Let $f(x) = \cos 2x + x$ on $[-\pi/2, \pi/2]$.
Greatest Value of \(f(x) = \cos(2x) + x\) on \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
Function Analysis: \(f(x) = \cos(2x) + x\)
The problem asks for the absolute maximum (greatest) value of the function \(f(x) = \cos(2x) + x\) over the specific interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This involves finding where the function reaches its highest point within these boundaries.
Derivative Calculation
To find potential maximum or minimum points (extrema), we first need to compute the derivative of the function \(f(x)\).
The derivative, \(f'(x)\), is found as follows:
\(f'(x) = \frac{d}{dx}(\cos(2x) + x)\)
Applying the chain rule to \(\cos(2x)\) and the power rule to \(x\):
\(f'(x) = -2\sin(2x) + 1\)
Critical Points Finding
Critical points are points in the domain where the derivative \(f'(x)\) is either zero or undefined. The derivative \(f'(x) = 1 - 2\sin(2x)\) is defined for all real numbers.
We set the derivative equal to zero to find the critical points:
\(1 - 2\sin(2x) = 0\)
\(2\sin(2x) = 1\)
\(\sin(2x) = \frac{1}{2}\)
We need to find the values of \(x\) within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This implies that \(2x\) must lie in the interval \([-\pi, \pi]\).
The angles \(\theta\) in the interval \([-\pi, \pi]\) for which \(\sin(\theta) = \frac{1}{2}\) are \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).
Therefore, we have:
- \(2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}\)
- \(2x = \frac{5\pi}{6} \implies x = \frac{5\pi}{12}\)
Both \(x = \frac{\pi}{12}\) and \(x = \frac{5\pi}{12}\) fall within our domain \([-\frac{\pi}{2}, \frac{\pi}{2}]\). These are the critical points we need to consider.
Evaluation at Key Points
To determine the greatest value of \(f(x)\) on the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\), we must evaluate the function at the critical points found (\(x = \frac{\pi}{12}\) and \(x = \frac{5\pi}{12}\)) and at the endpoints of the interval (\(x = -\frac{\pi}{2}\) and \(x = \frac{\pi}{2}\)).
| Point (\(x\)) | Function Evaluation (\(f(x)\)) | Resulting Value |
| \(x = -\frac{\pi}{2}\) | \(f(-\frac{\pi}{2}) = \cos(2 \times (-\frac{\pi}{2})) + (-\frac{\pi}{2}) = \cos(-\pi) - \frac{\pi}{2}\) | \(-1 - \frac{\pi}{2}\) |
| \(x = \frac{\pi}{12}\) | \(f(\frac{\pi}{12}) = \cos(2 \times \frac{\pi}{12}) + \frac{\pi}{12} = \cos(\frac{\pi}{6}) + \frac{\pi}{12}\) | \(\frac{\sqrt{3}}{2} + \frac{\pi}{12}\) |
| \(x = \frac{5\pi}{12}\) | \(f(\frac{5\pi}{12}) = \cos(2 \times \frac{5\pi}{12}) + \frac{5\pi}{12} = \cos(\frac{5\pi}{6}) + \frac{5\pi}{12}\) | \(-\frac{\sqrt{3}}{2} + \frac{5\pi}{12}\) |
| \(x = \frac{\pi}{2}\) | \(f(\frac{\pi}{2}) = \cos(2 \times \frac{\pi}{2}) + \frac{\pi}{2} = \cos(\pi) + \frac{\pi}{2}\) | \(-1 + \frac{\pi}{2}\) |
Comparing Function Values
Let's compare the numerical values we obtained:
- \(f(-\frac{\pi}{2}) = -1 - \frac{\pi}{2} \approx -1 - 1.571 = -2.571\)
- \(f(\frac{\pi}{12}) = \frac{\sqrt{3}}{2} + \frac{\pi}{12} \approx 0.866 + 0.262 = 1.128\)
- \(f(\frac{5\pi}{12}) = -\frac{\sqrt{3}}{2} + \frac{5\pi}{12} \approx -0.866 + 1.309 = 0.443\)
- \(f(\frac{\pi}{2}) = -1 + \frac{\pi}{2} \approx -1 + 1.571 = 0.571\)
By comparing these values, we can see which one is the largest.
Conclusion: Maximum Found
The largest value among \( -1 - \frac{\pi}{2}\), \(\frac{\sqrt{3}}{2} + \frac{\pi}{12}\), \(-\frac{\sqrt{3}}{2} + \frac{5\pi}{12}\), and \(-1 + \frac{\pi}{2}\) is \(\frac{\sqrt{3}}{2} + \frac{\pi}{12}\).
Therefore, the greatest value of the function \(f(x) = \cos(2x) + x\) on the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) is \(\frac{\sqrt{3}}{2} + \frac{\pi}{12}\).
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