Continuity Condition: Finding 'a' and 'b'

The function $f(x)$ is given as:

$f(x) = \begin{cases} \frac{ax^2 + 2ax + 3}{4x^2 + 4x - 3} & , x \neq -\frac{3}{2}, \frac{1}{2} \\ b & , x = -\frac{3}{2}, \frac{1}{2} \end{cases}$

For continuity at $x = -\frac{3}{2}$, the limit must equal the function value:

$ \lim_{x \to -\frac{3}{2}} f(x) = f(-\frac{3}{2}) = b $

The denominator is $4x^2 + 4x - 3 = (2x-1)(2x+3)$. As $x \to -\frac{3}{2}$, the denominator approaches 0.

For the limit to exist, the numerator must also approach 0 at $x = -\frac{3}{2}$:

$ a\left(-\frac{3}{2}\right)^2 + 2a\left(-\frac{3}{2}\right) + 3 = 0 $

$ \frac{9a}{4} - 3a + 3 = 0 $

$ -\frac{3a}{4} = -3 \implies a = 4 $

Substitute $a=4$ into the numerator: $4x^2 + 8x + 3 = (2x+1)(2x+3)$.

Now, calculate the limit:

$ \lim_{x \to -\frac{3}{2}} \frac{(2x+1)(2x+3)}{(2x-1)(2x+3)} = \lim_{x \to -\frac{3}{2}} \frac{2x+1}{2x-1} $

$ = \frac{2(-\frac{3}{2})+1}{2(-\frac{3}{2})-1} = \frac{-3+1}{-3-1} = \frac{-2}{-4} = \frac{1}{2} $

Thus, $b = \frac{1}{2}$.

Simplified Function Definition

With $a=4$, the function simplifies to:

$f(x) = \frac{2x+1}{2x-1}$ for $x \neq -\frac{3}{2}, \frac{1}{2}$.

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

Solving $f(f(x)) = \frac{7}{5}$

Let $y = f(x)$. We need to find $f(y) = \frac{7}{5}$.

First, find the expression for $f(f(x))$:

$ f(f(x)) = f\left(\frac{2x+1}{2x-1}\right) $

Assuming $f(x) \neq \frac{1}{2}$ (which implies $x \neq -\frac{3}{2}$), we use the simplified form:

$ f(f(x)) = \frac{2\left(\frac{2x+1}{2x-1}\right)+1}{2\left(\frac{2x+1}{2x-1}\right)-1} = \frac{\frac{2(2x+1)+(2x-1)}{2x-1}}{\frac{2(2x+1)-(2x-1)}{2x-1}} $

$ = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3} $

Now set this equal to $\frac{7}{5}$:

$ \frac{6x+1}{2x+3} = \frac{7}{5} $

Cross-multiply:

$ 5(6x+1) = 7(2x+3) $

$ 30x + 5 = 14x + 21 $

$ 16x = 16 $

$ x = 1 $

Check validity:

• $x=1$ is allowed ($1 \neq -\frac{3}{2}, \frac{1}{2}$).
• $f(1) = \frac{2(1)+1}{2(1)-1} = 3$. This value $3$ is not $-\frac{3}{2}$ or $\frac{1}{2}$.
• $f(f(1)) = f(3) = \frac{2(3)+1}{2(3)-1} = \frac{7}{5}$. This matches the requirement.

Therefore, the solution is $x = 1$.