The minimum value of function `f(x)=3x^4-8x^3+12x^2-48x+25` on [0,3] is equal to

Options

A. 25
B. -39
C. -25
D. 39

Answer

To find the minimum value of the function \( f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25 \) on the interval \([0, 3]\), we will follow these steps:

### Step 1: Find the derivative of the function
We start by finding the first derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(3x^4 - 8x^3 + 12x^2 - 48x + 25) \]
Calculating the derivative:
\[ f'(x) = 12x^3 - 24x^2 + 24x - 48 \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the first derivative equal to zero to find the critical points:
\[ 12x^3 - 24x^2 + 24x - 48 = 0 \]
Dividing the entire equation by 12:
\[ x^3 - 2x^2 + 2x - 4 = 0 \]

### Step 3: Factor the cubic equation
We can try to factor the cubic equation. We can use the Rational Root Theorem or synthetic division to find possible roots. Testing \( x = 2 \):
\[ 2^3 - 2(2^2) + 2(2) - 4 = 8 - 8 + 4 - 4 = 0 \]
Thus, \( x = 2 \) is a root. Now we can factor \( x - 2 \) out:
\[ x^3 - 2x^2 + 2x - 4 = (x - 2)(x^2 + 2) \]
The quadratic \( x^2 + 2 \) has no real roots since it is always positive.

### Step 4: Identify critical points
The only critical point in the interval \([0, 3]\) is \( x = 2 \).

### Step 5: Evaluate the function at the critical point and endpoints
Now we need to evaluate \( f(x) \) at the critical point and the endpoints of the interval:
1. At \( x = 0 \):
\[ f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25 \]
2. At \( x = 2 \):
\[ f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25 \]
\[ = 3(16) - 8(8) + 12(4) - 48(2) + 25 \]
\[ = 48 - 64 + 48 - 96 + 25 = -39 \]
3. At \( x = 3 \):
\[ f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25 \]
\[ = 3(81) - 8(27) + 12(9) - 48(3) + 25 \]
\[ = 243 - 216 + 108 - 144 + 25 = -12 \]

### Step 6: Compare values to find the minimum
Now we compare the values:
- \( f(0) = 25 \)
- \( f(2) = -39 \)
- \( f(3) = -12 \)

The minimum value of \( f(x) \) on the interval \([0, 3]\) is:
\[ \text{Minimum value} = -39 \]

### Final Answer
The minimum value of the function \( f(x) \) on the interval \([0, 3]\) is \(-39\).

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