Show that the maximum and minimum values of the function `(x+1)^2/(x+3)^3` are respectively given by `2/27 and 0.`

Answer

To find the maximum and minimum values of the function \( f(x) = \frac{(x+1)^2}{(x+3)^3} \), we will follow these steps:

### Step 1: Find the derivative using the quotient rule
The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

In our case, let \( g(x) = (x+1)^2 \) and \( h(x) = (x+3)^3 \).

- First, we find \( g'(x) \):
\[ g'(x) = 2(x+1) \]

- Next, we find \( h'(x) \):
\[ h'(x) = 3(x+3)^2 \]

Now, applying the quotient rule:
\[ f'(x) = \frac{2(x+1)(x+3)^3 - (x+1)^2 \cdot 3(x+3)^2}{(x+3)^6} \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ 2(x+1)(x+3)^3 - 3(x+1)^2(x+3)^2 = 0 \]

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

This gives us two factors:
1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( (x+3)^2 = 0 \) → \( x = -3 \) (not defined for \( f(x) \) since it leads to division by zero)
3. Solving \( 2(x+3) - 3(x+1) = 0 \):
\[ 2x + 6 - 3x - 3 = 0 \implies -x + 3 = 0 \implies x = 3 \]

### Step 3: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points \( x = -1 \) and \( x = 3 \):

- For \( x = -1 \):
\[ f(-1) = \frac{(-1+1)^2}{(-1+3)^3} = \frac{0^2}{2^3} = 0 \]

- For \( x = 3 \):
\[ f(3) = \frac{(3+1)^2}{(3+3)^3} = \frac{4^2}{6^3} = \frac{16}{216} = \frac{2}{27} \]

### Step 4: Determine maximum and minimum values
From the evaluations:
- The maximum value of \( f(x) \) is \( \frac{2}{27} \) at \( x = 3 \).
- The minimum value of \( f(x) \) is \( 0 \) at \( x = -1 \).

### Conclusion
Thus, we have shown that the maximum and minimum values of the function \( f(x) = \frac{(x+1)^2}{(x+3)^3} \) are \( \frac{2}{27} \) and \( 0 \), respectively.

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