find the coordinates of all extreme values (absolute and local) of f(x)=x-10 root x
.
Enter each extremum as an ordered pair, e.g., (2,3); if there is more than one extremum of a given type, enter a comma-separated list of ordered pairs. If there are no extrema of a given type, enter "none".

Absolute maxima: 
Absolute minima: 
Relative maxima: 
Relative minima

To find the extrema of the function \( f(x) = x - 10\sqrt{x} \), we need to:

1. Find the first derivative \( f'(x) \) and solve for the critical points.
2. Classify the critical points as relative maxima, minima, or neither using the first or second derivative test.
3. Check the behavior at the boundaries of the domain for absolute extrema.

### Step 1: Find the derivative of the function
The function is \( f(x) = x - 10\sqrt{x} \), which we can rewrite as \( f(x) = x - 10x^{1/2} \).

Now, take the derivative:
\[
f'(x) = 1 - 10 \cdot \frac{1}{2} x^{-1/2}
      = 1 - \frac{5}{\sqrt{x}}
\]

### Step 2: Find critical points
Set \( f'(x) = 0 \) to find critical points:
\[
1 - \frac{5}{\sqrt{x}} = 0
\]
Solving for \( x \):
\[
\frac{5}{\sqrt{x}} = 1 \quad \Rightarrow \quad \sqrt{x} = 5 \quad \Rightarrow \quad x = 25
\]

So, \( x = 25 \) is a critical point.

### Step 3: Classify the critical points
We will use the second derivative test to classify the critical point.

The second derivative of \( f(x) \) is:
\[
f''(x) = \frac{d}{dx}\left( 1 - \frac{5}{\sqrt{x}} \right)
       = 0 + 5 \cdot \frac{1}{2} x^{-3/2}
       = \frac{5}{2x^{3/2}}
\]

At \( x = 25 \), the second derivative is:
\[
f''(25) = \frac{5}{2 \cdot 25^{3/2}} = \frac{5}{2 \cdot 125} = \frac{5}{250} = \frac{1}{50}
\]
Since \( f''(25) > 0 \), the function is concave up at \( x = 25 \), indicating a **local minimum** at \( x = 25 \).

### Step 4: Evaluate endpoints for absolute extrema
The domain of the function is \( x \geq 0 \), because of the square root term.

- At \( x = 0 \):
\[
f(0) = 0 - 10\sqrt{0} = 0
\]
- At \( x = 25 \):
\[
f(25) = 25 - 10\sqrt{25} = 25 - 50 = -25
\]

Now check the behavior as \( x \to \infty \):
\[
\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( x - 10\sqrt{x} \right) = \infty
\]
As \( x \to \infty \), the function grows without bound, so there is no absolute maximum.

### Step 5: Conclusion
- **Absolute maxima:** None
- **Absolute minima:** \( (25, -25) \)
- **Relative maxima:** None
- **Relative minima:** \( (25, -25) \)

Do you want further details or have any questions?

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Here are 5 related questions:
1. How does the first derivative test help identify extrema?
2. What is the difference between absolute and relative extrema?
3. How do we determine the domain of a function involving square roots?
4. How do the second derivative and concavity relate to identifying local minima and maxima?
5. What is the significance of behavior at infinity for absolute extrema?

**Tip:** Always verify whether endpoints of a function's domain can yield absolute extrema, especially when the domain is limited.

Find the coordinates of all extreme values (absolute and local) of f(x) = x - 10√x. Enter each extremum as an ordered pair, e.g., (2,3); if there is more than one extremum of a given type, enter a comma-separated list of ordered pairs. If there are no extrema of a given type, enter 'none'.

Learn how to find the coordinates of extreme values of the function f(x) = x - 10√x. This detailed solution involves finding the first and second derivatives to determine relative and absolute minima and maxima. Suitable for students studying undergraduate calculus.

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