Let `f(x)=(x)/(x+3)`, then `(1)/(f(x+1))-f((1)/(x+1))=?`

Options

A. `(3x^(2)+7x+4)/(3(x^(2)+5x+5))`
B. `(3(x^(2)+5x+5))/(3x^(2)+7x+4)`
C. `(-x(2x+3))/((x+3)^(2))`
D. `(-(x+3)^(2))/(x(2x+3))`

Answer

To solve the problem \( \frac{1}{f(x+1)} - f\left(\frac{1}{x+1}\right) \) where \( f(x) = \frac{x}{x+3} \), we will follow these steps:

### Step 1: Calculate \( f(x+1) \)
We start by substituting \( x+1 \) into the function \( f(x) \):
\[ f(x+1) = \frac{x+1}{(x+1)+3} = \frac{x+1}{x+4} \]

### Step 2: Calculate \( \frac{1}{f(x+1)} \)
Now we find the reciprocal of \( f(x+1) \):
\[ \frac{1}{f(x+1)} = \frac{1}{\frac{x+1}{x+4}} = \frac{x+4}{x+1} \]

### Step 3: Calculate \( f\left(\frac{1}{x+1}\right) \)
Next, we substitute \( \frac{1}{x+1} \) into the function \( f(x) \):
\[ f\left(\frac{1}{x+1}\right) = \frac{\frac{1}{x+1}}{\frac{1}{x+1}+3} = \frac{\frac{1}{x+1}}{\frac{1 + 3(x+1)}{x+1}} = \frac{1}{1 + 3(x+1)} = \frac{1}{3x + 4} \]

### Step 4: Substitute into the expression
Now we substitute the results from Steps 2 and 3 into the original expression:
\[ \frac{1}{f(x+1)} - f\left(\frac{1}{x+1}\right) = \frac{x+4}{x+1} - \frac{1}{3x + 4} \]

### Step 5: Find a common denominator
The common denominator for the two fractions is \((x+1)(3x+4)\):
\[ \frac{x+4}{x+1} = \frac{(x+4)(3x+4)}{(x+1)(3x+4)} \]
\[ \frac{1}{3x + 4} = \frac{(x+1)}{(x+1)(3x+4)} \]

### Step 6: Combine the fractions
Now we can combine the fractions:
\[ \frac{(x+4)(3x+4) - (x+1)}{(x+1)(3x+4)} \]

### Step 7: Expand the numerator
Expanding the numerator:
\[ (x+4)(3x+4) = 3x^2 + 4x + 12x + 16 = 3x^2 + 16x + 16 \]
So, the numerator becomes:
\[ 3x^2 + 16x + 16 - (x + 1) = 3x^2 + 16x + 16 - x - 1 = 3x^2 + 15x + 15 \]

### Step 8: Final expression
Thus, the final expression is:
\[ \frac{3x^2 + 15x + 15}{(x+1)(3x+4)} \]

### Conclusion
The final answer is:
\[ \frac{3(x^2 + 5x + 5)}{(x+1)(3x+4)} \]