If `f(x) = x^5 - 5 x^4 + 5x^3 -10` has local max. and min. at x = p and x = q resp. , then (p,q) =

Options

A. (0,1)
B. (1,3)
C. (1,0)
D. None of these

Answer

To find the local maxima and minima of the function \( f(x) = x^5 - 5x^4 + 5x^3 - 10 \), we will follow these steps:

### Step 1: Find the first derivative
We need to differentiate the function to find the critical points where the local maxima and minima occur.

\[ f'(x) = \frac{d}{dx}(x^5 - 5x^4 + 5x^3 - 10) \]

Calculating the derivative:

\[ f'(x) = 5x^4 - 20x^3 + 15x^2 \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ 5x^4 - 20x^3 + 15x^2 = 0 \]

Factoring out the common term:

\[ 5x^2(x^2 - 4x + 3) = 0 \]

### Step 3: Solve for critical points
Now we can solve for \( x \):

1. From \( 5x^2 = 0 \), we get \( x = 0 \).
2. From \( x^2 - 4x + 3 = 0 \), we can factor it as:

\[ (x - 1)(x - 3) = 0 \]

Thus, we have \( x = 1 \) and \( x = 3 \).

The critical points are \( x = 0, 1, 3 \).

### Step 4: Find the second derivative
Next, we need to find the second derivative to determine the nature of these critical points:

\[ f''(x) = \frac{d}{dx}(5x^4 - 20x^3 + 15x^2) \]

Calculating the second derivative:

\[ f''(x) = 20x^3 - 60x^2 + 30x \]

### Step 5: Evaluate the second derivative at critical points
Now we will evaluate \( f''(x) \) at the critical points:

1. For \( x = 0 \):

\[ f''(0) = 20(0)^3 - 60(0)^2 + 30(0) = 0 \]

2. For \( x = 1 \):

\[ f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 \]

3. For \( x = 3 \):

\[ f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 20(27) - 60(9) + 90 = 540 - 540 + 90 = 90 \]

### Step 6: Determine maxima and minima
- Since \( f''(1) < 0 \), \( x = 1 \) is a local maximum.
- Since \( f''(3) > 0 \), \( x = 3 \) is a local minimum.
- The second derivative at \( x = 0 \) is zero, so we cannot conclude anything about it from the second derivative test.

### Conclusion
Thus, the local maximum occurs at \( x = 1 \) (denote as \( p \)) and the local minimum occurs at \( x = 3 \) (denote as \( q \)). Therefore, the answer is:

\[ (p, q) = (1, 3) \]