Let f(x) `=(40)/(3x^(4) +8x^(3) -18x^(2)+60).` Which of the following statement (s) about f(x) is (are ) correct ?

Options

A. f(x) has local minima at x=0
B. f(x) has local maxima at x=0
C. absolute maximum value of f(x) is not defined .
D. f(x) is local maxima at x=- 3, x =1

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{40}{3x^4 + 8x^3 - 18x^2 + 60} \) and determine the statements about it. We will find the critical points by differentiating the function and then analyze the behavior of the function at those points.

### Step 1: Differentiate the function
We will use the quotient rule to differentiate \( f(x) \).

The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then:
\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

Here, \( g(x) = 40 \) and \( h(x) = 3x^4 + 8x^3 - 18x^2 + 60 \).

Calculating \( g'(x) \):
\[ g'(x) = 0 \]

Calculating \( h'(x) \):
\[ h'(x) = 12x^3 + 24x^2 - 36x \]

Now applying the quotient rule:
\[ f'(x) = \frac{0 \cdot h(x) - 40 \cdot h'(x)}{(h(x))^2} = \frac{-40(12x^3 + 24x^2 - 36x)}{(3x^4 + 8x^3 - 18x^2 + 60)^2} \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ -40(12x^3 + 24x^2 - 36x) = 0 \]
This simplifies to:
\[ 12x^3 + 24x^2 - 36x = 0 \]
Factoring out \( 12x \):
\[ 12x(x^2 + 2x - 3) = 0 \]

### Step 3: Solve for critical points
Setting each factor to zero gives:
1. \( 12x = 0 \) → \( x = 0 \)
2. \( x^2 + 2x - 3 = 0 \)

Factoring the quadratic:
\[ (x + 3)(x - 1) = 0 \]
Thus, the critical points are:
\[ x = -3, \quad x = 0, \quad x = 1 \]

### Step 4: Analyze the critical points
To determine whether these points are maxima or minima, we can use the first derivative test. We will check the sign of \( f'(x) \) around the critical points.

- For \( x < -3 \), choose \( x = -4 \):
- \( f'(-4) > 0 \) (positive)

- For \( -3 < x < 0 \), choose \( x = -1 \):
- \( f'(-1) < 0 \) (negative)

- For \( 0 < x < 1 \), choose \( x = 0.5 \):
- \( f'(0.5) > 0 \) (positive)

- For \( x > 1 \), choose \( x = 2 \):
- \( f'(2) < 0 \) (negative)

### Step 5: Identify local maxima and minima
From the sign changes:
- At \( x = -3 \): Local maximum
- At \( x = 0 \): Local minimum
- At \( x = 1 \): Local maximum

### Step 6: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points:
- \( f(-3) = \frac{40}{3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60} \)
- \( f(0) = \frac{40}{60} = \frac{2}{3} \)
- \( f(1) = \frac{40}{3(1)^4 + 8(1)^3 - 18(1)^2 + 60} \)

Calculating \( f(-3) \) and \( f(1) \) will yield the same value, indicating that there is no absolute maximum.

### Conclusion
The statements about the function \( f(x) \) are:
1. Local minima at \( x = 0 \)
2. Local maxima at \( x = -3 \) and \( x = 1 \)
3. No absolute maximum exists.