If ` x in R , " then " f(x)=(x-1)^(2)+(x-2)^2+(x-3)^(2)+(x-4)^(2)` assumes its minimum value at :

Options

A. x=10
B. x=2.5
C. x=20
D. None of these

Answer

To find the minimum value of the function \( f(x) = (x-1)^2 + (x-2)^2 + (x-3)^2 + (x-4)^2 \), we will follow these steps:

### Step 1: Write the function
We start with the given function:
\[ f(x) = (x-1)^2 + (x-2)^2 + (x-3)^2 + (x-4)^2 \]

### Step 2: Differentiate the function
To find the minimum value, we need to find the first derivative \( f'(x) \):
\[ f'(x) = 2(x-1) + 2(x-2) + 2(x-3) + 2(x-4) \]
This simplifies to:
\[ f'(x) = 2[(x-1) + (x-2) + (x-3) + (x-4)] \]

### Step 3: Simplify the derivative
Now we simplify the expression inside the brackets:
\[ f'(x) = 2[4x - (1 + 2 + 3 + 4)] \]
Calculating the sum:
\[ 1 + 2 + 3 + 4 = 10 \]
Thus, we have:
\[ f'(x) = 2(4x - 10) = 8x - 20 \]

### Step 4: Set the derivative to zero
To find the critical points, we set the first derivative equal to zero:
\[ 8x - 20 = 0 \]

### Step 5: Solve for \( x \)
Solving for \( x \):
\[ 8x = 20 \implies x = \frac{20}{8} = 2.5 \]

### Step 6: Verify if it's a minimum
To confirm that this point is a minimum, we need to find the second derivative \( f''(x) \):
\[ f''(x) = 8 \]
Since \( f''(x) = 8 > 0 \), this indicates that \( f(x) \) is concave up at \( x = 2.5 \), confirming that it is indeed a minimum.

### Conclusion
The function \( f(x) \) assumes its minimum value at:
\[ \boxed{2.5} \]