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, we need to evaluate the expression \( \frac{1}{f(x+1)} - f\left(\frac{1}{x+1}\right) \) where \( f(x) = \frac{x}{x+3} \).

### 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 \( \frac{1}{f(x+1)} \)

Next, we find the reciprocal of \( f(x+1) \):

\[ \frac{1}{f(x+1)} = \frac{x+4}{x+1} \]

### Step 3: Calculate \( f\left(\frac{1}{x+1}\right) \)

Now, we need to find \( f\left(\frac{1}{x+1}\right) \):

\[ f\left(\frac{1}{x+1}\right) = \frac{\frac{1}{x+1}}{\frac{1}{x+1} + 3} = \frac{\frac{1}{x+1}}{\frac{1 + 3(x+1)}{x+1}} = \frac{1}{1 + 3(x+1)} = \frac{1}{3x + 4} \]

### Step 4: Substitute into the expression

Now, we substitute \( \frac{1}{f(x+1)} \) and \( f\left(\frac{1}{x+1}\right) \) into the 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

To combine these fractions, we need a common denominator, which is \( (x+1)(3x+4) \):

\[ \frac{(x+4)(3x+4) - (x+1)}{(x+1)(3x+4)} \]

### Step 6: Simplify the numerator

Now we simplify the numerator:

1. Expand \( (x+4)(3x+4) \):
\[ = 3x^2 + 4x + 12x + 16 = 3x^2 + 16x + 16 \]

2. Combine with \( -(x+1) \):
\[ = 3x^2 + 16x + 16 - x - 1 = 3x^2 + 15x + 15 \]

### Step 7: Final expression

Thus, we have:

\[ \frac{3x^2 + 15x + 15}{(x+1)(3x+4)} \]

### Step 8: Factor the numerator

We can factor the numerator:

\[ 3(x^2 + 5x + 5) \]

So the final answer is:

\[ \frac{3(x^2 + 5x + 5)}{(x+1)(3x+4)} \]

### Final Answer:

\[ \frac{3(x^2 + 5x + 5)}{(x+1)(3x+4)} \]

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