How to show that expression $$\frac{px^{2}+3x-4}{p+3x-4x^2}$$ will be capable of all values when $x$ is real,provided that $p$ has any value between $1$ and $7$?
Regarding my personal attempts,they are all futile.
I tried to expand $y=(px^{2}+3x-4)/(p+3x-4x^2$) to get a relation between $y$ and $p$ using "$b^{2}-4ac\geq0$"formula and substitute y with $1$.But,all futile.
Satrting from $$y=\frac{px^{2}+3x-4}{p+3x-4x^2}$$ rewrite the equation as a quadratic in $x$; this gives $$ (p+4 y)x^2-3 (y-1)x-(4+py)=0$$ Compute the discriminant $$\Delta_x=(16 p+9) y^2+\left(4 p^2+46\right) y+16 p+9$$ Compute again the discriminant for this quadratic in $y$ $$\Delta_y=p^4-41 p^2-72 p+112$$ and this one must be negative or zero.
By inspection and succesive divisions, you could notice that this factors $$\Delta_y=(p-7) (p-1) (p+4)^2$$
I am sure that you can take from here.