The point (s) of local maxima and local minima of the funcation f(x) `= 3x^(4) -8x^(3) +12 x^(2)-48 x +1 ` are :

Options

A. 2
B. `pm 2`
C. ` 2 , - sqrt(2)`
D. `- sqrt(2)`

Answer

To find the points of local maxima and minima of the function \( f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 1 \), we will follow these steps:

### Step 1: Find the first derivative of the function
We start by differentiating \( f(x) \) with respect to \( x \).

\[ f'(x) = \frac{d}{dx}(3x^4 - 8x^3 + 12x^2 - 48x + 1) \]

Using the power rule of differentiation:

\[ f'(x) = 12x^3 - 24x^2 + 24 - 48 \]

Simplifying this gives:

\[ f'(x) = 12x^3 - 24x^2 - 24 \]

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

\[ 12x^3 - 24x^2 - 24 = 0 \]

Dividing the entire equation by 12 simplifies it:

\[ x^3 - 2x^2 - 2 = 0 \]

### Step 3: Solve the cubic equation
Now, we need to solve the cubic equation \( x^3 - 2x^2 - 2 = 0 \). We can try to find rational roots using the Rational Root Theorem or by checking some simple values.

Testing \( x = 2 \):

\[ 2^3 - 2(2^2) - 2 = 8 - 8 - 2 = -2 \quad \text{(not a root)} \]

Testing \( x = -1 \):

\[ (-1)^3 - 2(-1)^2 - 2 = -1 - 2 - 2 = -5 \quad \text{(not a root)} \]

Testing \( x = 2 \) again:

\[ 2^3 - 2(2^2) - 2 = 8 - 8 - 2 = -2 \quad \text{(not a root)} \]

Testing \( x = -2 \):

\[ (-2)^3 - 2(-2)^2 - 2 = -8 - 8 - 2 = -18 \quad \text{(not a root)} \]

Testing \( x = 4 \):

\[ 4^3 - 2(4^2) - 2 = 64 - 32 - 2 = 30 \quad \text{(not a root)} \]

After testing various values, we find that \( x = 2 \) is indeed a root.

### Step 4: Factor the cubic polynomial
Since \( x = 2 \) is a root, we can factor \( x - 2 \) out of the polynomial. We can perform synthetic division or polynomial long division to find the other factors.

Using synthetic division with \( x = 2 \):

\[ \begin{array}{r|rrrr} 2 & 1 & -2 & 0 & -2 \\ & & 2 & 0 & 0 \\ \hline & 1 & 0 & 0 & -2 \\ \end{array} \]

This gives us:

\[ x^3 - 2x^2 - 2 = (x - 2)(x^2 + 0x + 1) = (x - 2)(x^2 + 1) \]

### Step 5: Find the critical points
Setting each factor to zero gives us:

1. \( x - 2 = 0 \) → \( x = 2 \)
2. \( x^2 + 1 = 0 \) → \( x^2 = -1 \) (no real solutions)

Thus, the only critical point is \( x = 2 \).

### Step 6: Determine if it is a maxima or minima
To determine whether this critical point is a local maximum or minimum, we can use the second derivative test.

### Step 7: Find the second derivative
We differentiate \( f'(x) \) again:

\[ f''(x) = \frac{d}{dx}(12x^3 - 24x^2 - 24) = 36x^2 - 48 \]

### Step 8: Evaluate the second derivative at the critical point
Now, we evaluate \( f''(2) \):

\[ f''(2) = 36(2^2) - 48 = 36(4) - 48 = 144 - 48 = 96 \]

Since \( f''(2) > 0 \), this indicates that \( x = 2 \) is a point of local minima.

### Conclusion
The point of local minima of the function \( f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 1 \) is at \( x = 2 \).

---