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The problem provided is about finding the minimum and maximum values of the directional derivative of a differentiable function \( f(x, y, z) \) at a given point.

Here's the breakdown:

### Problem 6:
You are given the gradient of the function \( f(x, y, z) \) at three points:

1. \( \nabla f(1,1,2) = 3i - j + 5k \)
2. \( \nabla f(2,1,1) = 6i - 2j + k \)
3. \( \nabla f(1,2,4) = i - j + k \)

Also, the variables are defined as:
\[
x = \frac{r+3s}{2}, \quad y = \frac{r+s}{2}, \quad z = r, \quad w = f(x, y, z)
\]
#### Part a:
The task is to **find the minimum and maximum values of the directional derivative** of \( f(x, y, z) \) at the point \( (1,1,2) \).

### Approach:
The directional derivative \( D_u f \) of \( f \) in the direction of a unit vector \( \mathbf{u} \) is given by:
\[
D_u f = \nabla f \cdot \mathbf{u}
\]
where \( \nabla f \) is the gradient of the function and \( \mathbf{u} \) is a unit vector.

1. The gradient at \( (1,1,2) \) is \( \nabla f(1,1,2) = 3i - j + 5k \).

2. The magnitude of this gradient vector is:
   \[
   |\nabla f(1,1,2)| = \sqrt{3^2 + (-1)^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35}
   \]

3. **Maximum value of the directional derivative**:
   - The maximum value occurs when \( \mathbf{u} \) is in the same direction as \( \nabla f \). So, the maximum value is the magnitude of the gradient:
   \[
   D_{\max} = |\nabla f(1,1,2)| = \sqrt{35}
   \]

4. **Minimum value of the directional derivative**:
   - The minimum value occurs when \( \mathbf{u} \) is in the opposite direction of \( \nabla f \). This will be the negative of the maximum value:
   \[
   D_{\min} = -|\nabla f(1,1,2)| = -\sqrt{35}
   \]

Thus, the maximum and minimum values of the directional derivative of \( f(x, y, z) \) at \( (1,1,2) \) are \( \sqrt{35} \) and \( -\sqrt{35} \), respectively.

Would you like me to explain further or go into more details about gradients or directional derivatives?

### Follow-up questions:
1. What is the interpretation of the gradient vector geometrically?
2. How is the unit vector related to the direction in which we compute the directional derivative?
3. Can we find the directional derivative in a specific direction, such as along the x-axis?
4. How would the problem change if the function was not differentiable at the given point?
5. What is the significance of the magnitude of the gradient in terms of function behavior?

### Tip:
The gradient of a function points in the direction of the steepest ascent, and its magnitude indicates the rate of change in that direction.

Suppose that f(x, y, z) is a differentiable function and given that ∇f(1,1,2) = 3i - j + 5k, ∇f(2,1,1) = 6i - 2j + k, ∇f(1,2,4) = i - j + k. Let x = (r + 3s) / 2, y = (r + s) / 2, z = r, and w = f(x, y, z). a. Find the minimum and maximum values of the directional derivative of f(x, y, z) at the point (1,1,2).

Learn how to find the maximum and minimum values of the directional derivative of f(x, y, z) at the point (1,1,2). This problem involves understanding gradients, directional derivatives, and applying vector calculus concepts.

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