To find the absolute minimum of the function $f(x) = \sin x + \cos x$ on the interval $0 \leq x \leq \frac{\pi}{2}$, follow these steps:

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1. Analyze the function

The function is continuous on $[0, \frac{\pi}{2}]$, so it attains its absolute minimum and maximum on this interval.

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2. Find critical points

We differentiate $f(x)$ with respect to $x$: $f'(x) = \cos x - \sin x$

Set $f'(x) = 0$ to find critical points: $\cos x - \sin x = 0$ $\cos x = \sin x$

This happens when: $x = \frac{\pi}{4}$

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3. Evaluate the function at critical points and endpoints

Now, evaluate $f(x)$ at the endpoints $x = 0$ and $x = \frac{\pi}{2}$, and the critical point $x = \frac{\pi}{4}$.

• At $x = 0$: $f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$

• At $x = \frac{\pi}{2}$: $f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1$

• At $x = \frac{\pi}{4}$: $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

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4. Compare values

The function values are: $f(0) = 1, \quad f\left(\frac{\pi}{2}\right) = 1, \quad f\left(\frac{\pi}{4}\right) = \sqrt{2}$

Since $\sqrt{2} \approx 1.414$, the absolute minimum is: $f(0) = 1 \text{ or } f\left(\frac{\pi}{2}\right) = 1$

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Final Answer:

The absolute minimum value of $f(x) = \sin x + \cos x$ on $[0, \frac{\pi}{2}]$ is 1.

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Would you like a detailed graph or additional explanations?

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Related Questions:

1. How do we compute the derivative of trigonometric functions in other examples?
2. What are the properties of the sine and cosine functions that help analyze their behavior?
3. How does the method change if the interval is extended beyond $[0, \frac{\pi}{2}]$?
4. What is the geometric interpretation of $f(x) = \sin x + \cos x$?
5. How can we use a unit circle to verify the result?

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Tip:

When analyzing trigonometric functions, pay attention to symmetry and periodicity—they can often simplify solving problems!