I am trying to solve the below problem in pre-calculus, I give below the steps I followed , I am stuck as it gets very complicated , need help in solution.

A function $f : \mathbb R \to \mathbb R$ is defined by $$f(x) = \frac{kx^2+6x-8}{k+6x-8x^2}.$$ Find the intervals of values of $k$ such that $f$ is onto. The answer given in the book is $2 \leq k \leq14$.

I started by trying to find the range of the function and find the values of k such that the range is = $\mathbb R$ ( that is range = co domain , in this case co domain is $\mathbb R$ and hence has to prove that range = $\mathbb R$)

Let $$y = \frac{kx^2+6x-8}{k+6x-8x^2}$$ Rearranging and grouping gives $$(k+8y)x^2 + 6(1-y)x - (8+ky) = 0.$$ This is quadratic in $x$ and for $x$ to be real $det(x) \geq 0$, which requires $$36(1-y)^2+4(k+8y)(8+ky) \geq 0.$$

Rearranging and grouping this equation gives $$(36+32k)y^2+(4k^2+184)y+(36+32k) \geq 0.$$ Now I have to find the range of values of $k$ such that the above expression is $\geq 0$. Totally stuck here. Please help. Also, is there any other easier way of arriving at the range of $k$ ?

  • 1
    $\begingroup$ For each value of $k$ between $2$ and $14$ there is a point where the function is not defined. Since the domain is specified as $\mathbb R$ I would consider the answer wrong. Only for $k <-9/8$ the function is defined on the whole line. $\endgroup$ Aug 26, 2018 at 11:43
  • $\begingroup$ Please read this MathJax tutorial, which explains how to typeset mathematics on this site. $\endgroup$ Aug 26, 2018 at 11:47

1 Answer 1


Start with your result $$(36+32k)y^2+(4k^2+184)y+(36+32k) \geq 0$$ If we draw a graph about $z=(36+32k)y^2+(4k^2+184)y+(36+32k)$ which axes are $y$ and $z$. We want this be a upper-open parabola and not touch the points below the horizontal axis. so what we want is \begin{cases}36+32k>0\\\Delta = (4k^2+184)^2-4(36+32k)^2\leq 0\end{cases}

First solve the second inequality, using formula of the different of squares, we have \begin{align}(4k^2+184+72+64k)(4k^2+184-72-64k)&\leq 0\\(k^2+16k+64)(k^2-16k+28)&\leq 0\\(k+8)^2(k-2)(k-14)&\leq 0\end{align} Since $(k+8)^2$ will always stay non-negative, it will have only little influence in the inequality, then we just have to solve $$(k-2)(k-14)\leq 0$$ which give us $2\leq k\leq 14$. Now see that for all $k$ in this range, $36+32k$ will always stay positive. Hence the answer is $2\leq k\leq 14$.

  • $\begingroup$ Wow ..thanks very much ..I think I gave up towards the fag end of the solution ... $\endgroup$ Aug 26, 2018 at 12:13
  • $\begingroup$ for $k=2$ the function is not onto $\endgroup$
    – Lozenges
    Aug 26, 2018 at 13:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .