We are given the function $f(x) = \frac{ax - b}{x + a}$, and some conditions:

1. $f(0) = -\frac{3}{2}$
2. $f'(0) = \frac{1}{4}$

We are asked to find $f(1)$.

Step 1: Use $f(0) = -\frac{3}{2}$

We know: $f(0) = \frac{a(0) - b}{0 + a} = \frac{-b}{a} = -\frac{3}{2}$ Thus, we can solve for $b$ as follows: $\frac{-b}{a} = -\frac{3}{2} \quad \Rightarrow \quad b = \frac{3a}{2}$

Step 2: Use $f'(x)$ and $f'(0) = \frac{1}{4}$

Now, let's compute the derivative $f'(x)$ using the quotient rule:

$f'(x) = \frac{(x + a)(a) - (ax - b)(1)}{(x + a)^2} = \frac{a(x + a) - (ax - b)}{(x + a)^2}$ Simplifying the numerator: $f'(x) = \frac{a(x + a) - ax + b}{(x + a)^2} = \frac{ax + a^2 - ax + b}{(x + a)^2} = \frac{a^2 + b}{(x + a)^2}$ Thus, the derivative is: $f'(x) = \frac{a^2 + b}{(x + a)^2}$

Using $f'(0) = \frac{1}{4}$: $f'(0) = \frac{a^2 + b}{a^2} = \frac{a^2 + b}{a^2} = \frac{1}{4}$ Substitute $b = \frac{3a}{2}$: $\frac{a^2 + \frac{3a}{2}}{a^2} = \frac{1}{4}$ Multiply both sides by $a^2$: $a^2 + \frac{3a}{2} = \frac{a^2}{4}$ To simplify, multiply everything by 4: $4a^2 + 6a = a^2$ Now, rearrange terms: $4a^2 - a^2 + 6a = 0 \quad \Rightarrow \quad 3a^2 + 6a = 0$ Factor out $a$: $a(3a + 6) = 0$ Thus, $a = 0$ or $a = -2$.

Step 3: Determine $b$

If $a = -2$, substitute into $b = \frac{3a}{2}$: $b = \frac{3(-2)}{2} = -3$

Step 4: Find $f(1)$

Now, substitute $a = -2$ and $b = -3$ into the function $f(x) = \frac{ax - b}{x + a}$:

$f(1) = \frac{-2(1) - (-3)}{1 + (-2)} = \frac{-2 + 3}{1 - 2} = \frac{1}{-1} = -1$

Final Answer:

$f(1) = -1$

Would you like further details or clarification? Here are some related questions:

1. What is the meaning of the quotient rule?
2. How do you find the derivative of a quotient in general?
3. How can you verify if the values of $a$ and $b$ are correct?
4. What happens to the function as $x \to \infty$?
5. How can $f'(x)$ help in finding critical points?

Tip: When using the quotient rule, keep the numerator and denominator separate and simplify carefully before differentiating.