To find the maximum value of (f(x) = \sin x + \cos x + \sin x \cos x) for (x \in [0, \pi/2]), follow these steps:

Step 1: Substitute (t = \sin x + \cos x)

We know:
[t^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x]
Thus:
[ \sin x \cos x = \frac{t^2 - 1}{2}]

Step 2: Range of (t)

[t = \sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)]
For (x \in [0, \pi/2]), (x + \pi/4 \in [\pi/4, 3\pi/4]), so (\sin\left(x + \pi/4\right) \in [\frac{\sqrt{2}}{2}, 1]).
Hence:
[t \in [1, \sqrt{2}]]

Step 3: Rewrite (f(x)) in terms of (t)

[f(x) = t + \frac{t^2 - 1}{2} = \frac{1}{2}t^2 + t - \frac{1}{2}]
This is a quadratic function (g(t) = \frac{1}{2}t^2 + t - \frac{1}{2}), opening upward with vertex at (t = -1).

Since (t \in [1, \sqrt{2}]) (right of the vertex), (g(t)) is increasing here. The maximum occurs at (t = \sqrt{2}).

Step 4: Calculate the maximum value

[g(\sqrt{2}) = \frac{1}{2}(\sqrt{2})^2 + \sqrt{2} - \frac{1}{2} = \frac{1}{2}(2) + \sqrt{2} - \frac{1}{2} = \sqrt{2} + \frac{1}{2}]

Answer: (\boxed{\sqrt{2} + \frac{1}{2}}) or (\boxed{\frac{1 + 2\sqrt{2}}{2}})

The simplest form is (\boxed{\sqrt{2} + \frac{1}{2}}), but if we rationalize it: (\boxed{\frac{1 + 2\sqrt{2}}{2}}). Either is correct, but let's use the combined form:

(\boxed{\frac{1 + 2\sqrt{2}}{2}})

(\boxed{\frac{1 + 2\sqrt{2}}{2}})

作者声明:本文包含人工智能生成内容。