Let f(x)=casesax^2+2ax+34x^2+4x-3, & x -32,12[2mm] b, & x=-32,12cases be continuous at x=-…

Question

Let $f(x)=\begin{cases}\dfrac{ax^2+2ax+3}{4x^2+4x-3}, & x\neq -\dfrac32,\dfrac12\[2mm] b, & x=-\dfrac32,\dfrac12\end{cases}$ be continuous at $x=-\dfrac32$. If $f(f(x))=\dfrac75$, then $x$ is equal to:

Accepted Answer

$4x^2+4x-3=(2x-1)(2x+3)$, which vanishes at $x=-\frac32,\frac12$. For $f$ to be continuous at $x=-\frac32$, the numerator must also vanish there:
$$a\left(\frac94\right)-3a+3=0 \implies -\frac{3a}{4}+3=0 \implies a=4$$

Numerator becomes $4x^2+8x+3=(2x+1)(2x+3)$, so for $x\ne-\frac32,\frac12$:
$$f(x)=\frac{(2x+1)(2x+3)}{(2x-1)(2x+3)}=\frac{2x+1}{2x-1}$$

and $b=\lim_{x\to-3/2}f(x)=\dfrac{-3+1}{-3-1}=\dfrac12$.

Now solve $f(f(x))=\dfrac75$. Let $y=f(x)$:
$$\frac{2y+1}{2y-1}=\frac75 \implies 5(2y+1)=7(2y-1)\implies 10y+5=14y-7\implies y=3$$

Then $\dfrac{2x+1}{2x-1}=3 \implies 2x+1=6x-3 \implies x=1$.

Solutions · 1