quadratics
Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real.
I don't know how to proceed with this question. I have equated this equation with $y$ to obtain a quadratic equation: $(m+4y)x^2+(3-3y)x-(4+my)=0$. Now I have no idea as to how I can find the answer. A small hint will be helpful.
it looks easier to consider the reciprocal $$y = g(x) = \frac{a+x^2}{x+1} = x-1 + \frac {a+1}{x+1}$$ and look for the values $a$ so that the range of $g$ contains $[1, \infty).$
we will break the problem in into three cases: $a > -1, a = -1$ and $ a < -1.$
consider the case $a > -1.$ here we have the graph of $y = g,$ hyperbola with a local min $$ y = 2\sqrt{a+1} - 1 \text{ at } x = -1 + \sqrt{a+1}.$$ for the local min to be less than or equal to $1,$ it is necessary that $$-1 < a \le 0.\tag 1$$
case $a = -1$ we have $g = x-1.$ the range of $g$ certainly contains $[1, \infty)$
in case $a < -1,$ the range of $a$ is $(-\infty, \infty)$ which contains $[1, \infty)$
we can conclude that the constraint on $a$ is $$ a \le 0 \implies range\left(\frac{x+1}{x^2 + a}\right) \subseteq[0,1]. $$
Let $y = \frac{x^2-x+c}{x^2+x+2c} $ be a value taken by the given fraction. This gives
\begin{equation*}
x^2(y-1) +x(y+1) +c(2y-1) = 0
\end{equation*}
should have a real root. Thus
\begin{equation*}
(y+1)^2 -4c(y-1)(2y-1) \geq 0
\end{equation*}
and hence
\begin{equation*}
y^2(1-8c) + 2y(1+6c)+(1-4c) \geq 0
\end{equation*}
If the given expression takes all values, then the above should be true for all $y$. Thus we have
\begin{equation*}
(1+6c)^2 - (1-8c)(1-4c) \leq 0
\end{equation*}
and $8c \leq 1$. The above expression simplifies to $24c+4c^2 \leq 0$ and hence $-6 \leq c \leq 0$.
when $c=-6$, the graph is:
Clearly, $y=1$ is an asymptote and hence the expression misses the value 1. Again, the same phenomenon is observed when $c=0$. Thus the required range is $-6 < c < 0$
For negative values outside the range, the graph is 
and for positive values of $c$, the graph is 
and for values within the range, the graph is 
Best Answer
Momo's answer is correct except for the values $m=1$ and $m=7$.
Let $f$ be given by $$f(x)=\frac{mx^2+3x-4}{-4x^2+3x+m}\;\;$$ If $m = 1$, then \begin{align*} f(x)&=\frac{x^2+3x-4}{-4x^2+3x+1}\\[4pt] &=-\frac{x^2+3x-4}{4x^2-3x-1}\\[4pt] &=-\frac{(x-1)(x+4)}{(x-1)(4x+1)}\\[4pt] &=-\frac{x+4}{4x+1},\;\;x\ne 1\\[4pt] \end{align*} which misses the values$\;y=-1/4\;$and$\;y=-1$.
If $m = 7$, then \begin{align*} f(x)&=\frac{7x^2+3x-4}{-4x^2+3x+7}\\[4pt] &=-\frac{7x^2+3x-4}{4x^2-3x-7}\\[4pt] &=-\frac{(x+1)(7x-4)}{(x+1)(4x-7)}\\[4pt] &=-\frac{7x-4}{4x-7},\;\;x\ne -1\\[4pt] \end{align*} which misses the values$\;y=-7/4\;$and$\;y=-1$.
Hence the correct answer is $1 < m < 7$.