Find the open intervals on which the function  f(x)=x+10√1-x
  is increasing or decreasing. The safe points will be calculated from these intervals. If the function is never increasing or decreasing, provide an input of NA to your computer.

To find the intervals on which the function \( f(x) = x + 10\sqrt{1 - x} \) is increasing or decreasing, we need to compute its derivative and analyze the sign of the derivative.

### Step 1: Compute the derivative of the function

The given function is:
\[
f(x) = x + 10\sqrt{1 - x}
\]

To differentiate this, we'll use the sum and chain rule.

- The derivative of \( x \) is 1.
- The derivative of \( 10\sqrt{1 - x} \) is \( 10 \times \frac{1}{2}(1 - x)^{-1/2} \times (-1) \) (by applying the chain rule).

Thus, the derivative of the function \( f(x) \) is:

\[
f'(x) = 1 - \frac{10}{2\sqrt{1 - x}} = 1 - \frac{5}{\sqrt{1 - x}}
\]

### Step 2: Find the critical points

To find the critical points, we set the derivative equal to zero and solve for \( x \):

\[
1 - \frac{5}{\sqrt{1 - x}} = 0
\]

\[
\frac{5}{\sqrt{1 - x}} = 1
\]

\[
\sqrt{1 - x} = 5
\]

Square both sides:

\[
1 - x = 25
\]

\[
x = -24
\]

However, \( \sqrt{1 - x} \) is only valid when \( 1 - x \geq 0 \), which implies \( x \leq 1 \). Thus, the critical point \( x = -24 \) is outside the domain of the function, and we disregard it.

### Step 3: Determine the domain of the function

For the function to be real-valued, the square root term \( \sqrt{1 - x} \) must have a non-negative argument. Thus, we require:

\[
1 - x \geq 0
\]

\[
x \leq 1
\]

Therefore, the domain of \( f(x) \) is \( (-\infty, 1] \).

### Step 4: Analyze the sign of the derivative

Now, we analyze the sign of the derivative \( f'(x) = 1 - \frac{5}{\sqrt{1 - x}} \).

- When \( x = 1 \), the derivative is undefined because the denominator \( \sqrt{1 - x} \) becomes zero.
- For \( x < 1 \), we need to check if the derivative is positive or negative.

The derivative \( f'(x) = 1 - \frac{5}{\sqrt{1 - x}} \) will be positive if:

\[
1 > \frac{5}{\sqrt{1 - x}}
\]

This can only happen if \( \sqrt{1 - x} > 5 \), but this is not possible because \( \sqrt{1 - x} \leq 1 \) for \( x \leq 1 \).

Thus, \( f'(x) \) is always negative for \( x < 1 \).

### Step 5: Conclusion

Since the derivative is always negative for \( x < 1 \), the function is **decreasing** on the interval \( (-\infty, 1) \) and undefined at \( x = 1 \).

Thus, the function is **never increasing** and **always decreasing** on its domain.

---

Would you like more details on any part of the solution or have any further questions? Here are some related questions:

1. How do we apply the chain rule when differentiating functions involving square roots?
2. Why is the derivative undefined at \( x = 1 \)?
3. What happens to the behavior of the function as \( x \) approaches 1 from the left?
4. How can we find intervals of increase or decrease for more complex functions?
5. What are the implications of a function being always decreasing?

**Tip:** Always check the domain of the function when it involves square roots or logarithms to avoid invalid critical points.

Find the open intervals on which the function f(x) = x + 10√1 - x is increasing or decreasing. The safe points will be calculated from these intervals. If the function is never increasing or decreasing, provide an input of NA to your computer.

Explore how to find the intervals where the function f(x) = x + 10√(1 - x) is increasing or decreasing. This problem uses derivatives, the chain rule, and critical point analysis to determine the behavior of the function. Suitable for advanced high school or early college calculus students.

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