Let $f(x) = \begin{cases} \frac{ax^2 + 2ax + 3}{4x^2 + 4x - 3} & , x \neq -\frac{3}{2}, \frac{1}{2} \\ b & , x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at $x = -\frac{3}{2}$. If $f(f(x)) = \frac{7}{5}$, then $x$ is equal to :
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 $x = 1$.
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