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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.

Vidyanshu Mishra
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oshhh
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  • Hint : It is necessary (but probably not sufficient), that the denominator has two real roots – Peter Dec 19 '16 at 12:39
  • See here: http://mathhelpforum.com/pre-calculus/186618-finding-interval-m-quadratic-equation.html –  Dec 19 '16 at 12:41
  • @Rohan doesn't really help...it just declares the question to be wrong. – oshhh Dec 19 '16 at 12:43
  • I just wanted you to know it is the same question and that person has tried in the same way as you have done. –  Dec 19 '16 at 12:43

4 Answers4

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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 $1 < m < 7$.

quasi
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  • I was just about to edit my answer. +1 – Momo Dec 19 '16 at 15:39
  • @quasi,for m=6 also,the expression in denominator has real root ,which makes it 0,so thus all such cases should have been rejected,why only cases of 1,7 are rejected? – Dheeraj Gujrathi Sep 24 '21 at 03:37
  • @Dheeraj Gujrathi: For $1 < m < 7$, the real roots of the denominator correspond to vertical asymptotes on the graph of the equation $y=f(x)$. But for $1 < m < 7$, the function $f$ is already onto when restricted to the values of $x$ between the real roots of the denominator. Try plotting $f$ for the case $m=6$ and you'll see. – quasi Sep 24 '21 at 05:58
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Hint

$$f(x)=\frac{mx^2+3x-4}{-4x^2+3x+m}$$

The function $f$ must to be onto. It means that:

$$p=\frac{mx^2+3x-4}{-4x^2+3x+m} \Rightarrow (m+4p)x^2+3x(1-p)-4-pm=0$$

The above equation must have roots for any $p \in \Bbb R$. It means that:

$$\Delta=9(1-p)^2+4(m+4p)(4+pm)\geq 0$$

for any choice of $p$.

$$(9+16m)p^2+(4m^2-46)p+9+16m \geq0$$

Then we have to analyze that new quadratic equation. Once the above expression must be always non negative, then.

$$\Delta'=(4m^2-46)^2-4(9+16m)^2=(m^2-8m-16)(m^2+8m-7)\leq 0$$

Solving the above inequality we have the values of $m$.

Arnaldo
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  • I totally understand your answer. On solving the final inequation, we get $x\in\left[-4-\sqrt {23}, 4-4\sqrt 2\right]\cup\left[\sqrt {23}-4,4+4\sqrt 2\right]$. But this doesn't seem to agree with the answer $x\in(1,7)$ provided in my book as well as this answer. Is there anything else we're missing? – Vishnu Jul 16 '20 at 10:51
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I don't know whether it's the simplest way but I would do this:

1) Your expanded equation $(m+4y)x^2+(3-3y)x-(4+my)=0$ needs to have real roots for all $y$ so you need to have $\Delta\ge 0$ for all $y$

2) Your $\Delta$ is itself a quadratic inequality in $y$ with positive coefficient, so to have $\Delta\ge 0$ for all $y$ you need $\Delta_y\le 0$

Which gives you $1\le m\le 7$

Momo
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Hint: $$\frac{mx^2+3x-4}{-4x^2+3x+m} = y$$ The denominator shouldn't have any real roots. Thus, $ \delta_{\text{denom}} = 3^2+16m< 0\Rightarrow m<-\frac{9}{16}$

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    But y should be able to take any value...how do we ensure that? – oshhh Dec 19 '16 at 12:34
  • If $m$ is chosen in a way the denominator has no roots, the function only has a finite range – Peter Dec 19 '16 at 12:34
  • @OsheenSachdev It cannot have every value if the denominator is never $0$ – Peter Dec 19 '16 at 12:35
  • For all $ x,m\in \mathbb{R}$ , numerator and denominator belong to $\mathbb{R}$ . You only need to study for which values of $m$ there exists $x$ such that $ -4x^2+3x+m=0$ . For those values of $m$ the quotient is not defined for all $x\in\mathbb{R}$ . –  Dec 19 '16 at 12:36
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    @Rohan Not the domain should be $\mathbb R$, but the range – Peter Dec 19 '16 at 12:40
  • Okay... how do we ensure that there is a real value of x for all real values of y? I mean assuming y to be some no, say 10, how do we ensure that there is a real value of x that makes the expression equal to y? – oshhh Dec 19 '16 at 12:41
  • That condition does not guarantee that the function is onto. Take $m=-1$ as an example and plot the graph. – Arnaldo Dec 19 '16 at 14:00