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.

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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.