Find the maxima and minimum value of the function `y = 40/(3x^(4)+ 8x^(3)- 18x^(2) +60`.

Answer

To find the maxima and minima of the function \( y = \frac{40}{3x^4 + 8x^3 - 18x^2 + 60} \), we will follow these steps:

### Step 1: Differentiate the function

First, we need to differentiate the function with respect to \( x \). We will use the quotient rule for differentiation.

Let \( u = 40 \) and \( v = 3x^4 + 8x^3 - 18x^2 + 60 \).

Using the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Since \( \frac{du}{dx} = 0 \), we only need to find \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = 12x^3 + 24x^2 - 36x \]

Thus, the derivative becomes:
\[ \frac{dy}{dx} = \frac{(3x^4 + 8x^3 - 18x^2 + 60)(0) - 40(12x^3 + 24x^2 - 36x)}{(3x^4 + 8x^3 - 18x^2 + 60)^2} \]
\[ = -\frac{40(12x^3 + 24x^2 - 36x)}{(3x^4 + 8x^3 - 18x^2 + 60)^2} \]

### Step 2: Set the derivative equal to zero

To find critical points, we set the numerator equal to zero:
\[ 12x^3 + 24x^2 - 36x = 0 \]

Factoring out \( 12x \):
\[ 12x(x^2 + 2x - 3) = 0 \]
\[ 12x(x + 3)(x - 1) = 0 \]

Thus, the critical points are:
\[ x = 0, \quad x = -3, \quad x = 1 \]

### Step 3: Determine maxima and minima using the second derivative test

Next, we need to find the second derivative \( \frac{d^2y}{dx^2} \). For simplicity, we will evaluate the first derivative at the critical points.

1. **At \( x = 0 \)**:
\[ \frac{dy}{dx} = -\frac{40(12(0)^3 + 24(0)^2 - 36(0))}{(3(0)^4 + 8(0)^3 - 18(0)^2 + 60)^2} = 0 \]

2. **At \( x = -3 \)**:
\[ \frac{dy}{dx} = -\frac{40(12(-3)^3 + 24(-3)^2 - 36(-3))}{(3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60)^2} \]
Calculate \( \frac{d^2y}{dx^2} \) to determine if it's a maximum or minimum.

3. **At \( x = 1 \)**:
\[ \frac{dy}{dx} = -\frac{40(12(1)^3 + 24(1)^2 - 36(1))}{(3(1)^4 + 8(1)^3 - 18(1)^2 + 60)^2} \]
Similarly, calculate \( \frac{d^2y}{dx^2} \).

### Step 4: Calculate the function values at critical points

1. **At \( x = 0 \)**:
\[ y(0) = \frac{40}{3(0)^4 + 8(0)^3 - 18(0)^2 + 60} = \frac{40}{60} = \frac{2}{3} \]

2. **At \( x = -3 \)**:
\[ y(-3) = \frac{40}{3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60} \]
Calculate this value.

3. **At \( x = 1 \)**:
\[ y(1) = \frac{40}{3(1)^4 + 8(1)^3 - 18(1)^2 + 60} \]
Calculate this value.

### Step 5: Compare values to find maxima and minima

Finally, compare the values of \( y \) at \( x = 0, -3, 1 \) to determine the maximum and minimum values.