6 votes
Accepted

Find the set of values of $\alpha$ so that $f(x)=\dfrac{\alpha x^2+6x-8}{\alpha+6x-8x^2}$ is one one.

Your strategy is good, and your computations are mostly correct, with a minor correction: $(1)$ implies that $$\color{red}{\alpha} \in [2, 14], \tag{1}$$ not $x$. And this is a correct conclusion, ...
heropup's user avatar
  • 136k
3 votes
Accepted

Find number of solutions to the equation $\sin(6\sin x)=\frac{x}{6}$.

If $x$ is a root of the equation, so is $-x$. Furthermore, $x=0$ is a root. For these reasons, we only enumerate the number of positive roots. Since $\frac{x}{6}>1$ for $x>6$, for every positive ...
Mostafa Ayaz's user avatar
  • 32.2k
1 vote

Find the set of values of $\alpha$ so that $f(x)=\dfrac{\alpha x^2+6x-8}{\alpha+6x-8x^2}$ is one one.

For $g(x)=-6x+8x^2$ and $h(x)=x^{-1}(\alpha -g(x))$ we have $f(x)={h(x^{-1})\over h(x)}.$ Therefore $f(x)$ should be undefined at $x=1$ or at $x=-1,$ as otherwise $f(1)=f(-1)=1.$ Hence $h(x)$ ...
Ryszard Szwarc's user avatar
1 vote

Find the set of values of $\alpha$ so that $f(x)=\dfrac{\alpha x^2+6x-8}{\alpha+6x-8x^2}$ is one one.

Hint Notice that if $\pm 1$ are both in the domain of $f$, then $$f(\pm 1) = \frac{\alpha (\pm 1)^2 + 6(\pm 1) - 8}{\alpha + 6(\pm 1) - 8(\pm 1)^2} = \frac{\alpha - 8 \pm 6}{\alpha - 8 \pm 6} = 1 .$$ ...
Travis Willse's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible