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I am trying to find the range of the function $f(x)=\frac{2x-5}{x^2+x-2}$ without using the graph. I usually use two methods to find the range.

  • The first method is finding the domain of the inverse function, but it’s complicated here.
  • I will explain the second method with an example, say we want to find the range of $\sqrt{16-x^2}$, since the domain is $[-4,4]$, then $-4\le x\le 4$, and hence $0\le \sqrt{16-x^2}\le 4$, which implies that the range is $[0,4]$.

I cannot seem to apply any of those methods for my functions. Any help would be appreciated.

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  • $\begingroup$ Have you learned how to find maxima and minima using calculus techniques? $\endgroup$ Sep 16, 2023 at 15:22
  • $\begingroup$ @N.F.Taussig Yes $\endgroup$
    – Dima
    Sep 16, 2023 at 15:57
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    $\begingroup$ The other method I had in mind was finding the domain, determining the asymptotes, finding the critical points by setting the derivative of the function equal to zero, then using the First Derivative Test or Second Derivative Test to find the relative maxima and minima of the function. $\endgroup$ Sep 16, 2023 at 16:25
  • $\begingroup$ @N.F.Taussig Thanks a lot. $\endgroup$
    – Dima
    Sep 24, 2023 at 18:48

2 Answers 2

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Suppose that $$y = \frac{2x - 5}{x^2 + x - 2}$$ is in the range of $f$. Then \begin{align*} y(x^2 + x - 2) & = 2x - 5\\ yx^2 + yx - 2y & = 2x - 5\\ yx^2 + (y - 2)x - 2y + 5 & = 0 \end{align*} which is a quadratic equation in $x$ unless $y = 0$.

Observe that $y = 0$ when $x = 5/2$. If $y \neq 0$, then a real number $y$ is in the range of $f$ if and only if $x$ is a real root of the quadratic equation $yx^2 + (y - 2)x - 2y + 5 = 0$.

A quadratic equation $ax^2 + bx + c = 0$ with real coefficients $a, b, c$ such that $a \neq 0$ has real roots if and only if its discriminant $\Delta = b^2 - 4ac \geq 0$.

Thus, if $y \neq 0$, $y$ is in the range of $f$ if and only if \begin{align*} (y - 2)^2 - 4y(-2y + 5) & \geq 0\\ y^2 - 4y + 4 + 8y^2 - 20y & \geq 0\\ 9y^2 - 24y + 4 & \geq 0\\ y^2 - \frac{8}{3}y + \frac{4}{9} & \geq 0\\ y^2 - \frac{8}{3}y & \geq -\frac{4}{9}\\ y^2 - \frac{8}{3}y + \frac{16}{9} & \geq \frac{12}{9} && \text{(complete the square)}\\ \left(y - \frac{4}{3}\right)^2 & \geq \frac{12}{9}\\ \left|y - \frac{4}{3}\right| & \geq \frac{2\sqrt{3}}{3} && \text{(take the square root of each side of the inequality)} \end{align*} Thus, $y = 0$ or \begin{align*} y - \frac{4}{3} & \geq \frac{2\sqrt{3}}{3} & \text{or} & & y - \frac{4}{3} & \leq -\frac{2\sqrt{3}}{3}\\ y & \geq \frac{4 + 2\sqrt{3}}{3} & & & y & \leq \frac{4 - 2\sqrt{3}}{3} \end{align*} Since $\frac{4 - 2\sqrt{3}}{3} \geq 0$, the range of $f$ is $$\left(-\infty, \frac{4 - 2\sqrt{3}}{3}\right] \cup \left[\frac{4 + 2\sqrt{3}}{3}, \infty\right)$$

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  • $\begingroup$ +1 . Your answer goes beyond the calculations and explains everything in a way that appeals to everyone . $\endgroup$ Sep 16, 2023 at 15:50
  • $\begingroup$ Thanks for the detailed answer, appreciate it. $\endgroup$
    – Dima
    Sep 16, 2023 at 15:55
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Let $$k = \frac{2x-5}{x^2+x-2}$$

We have $$kx^2+(k-2)x-2k+5=0.$$

We want the discriminant to be nonnegative.

$$(k-2)^2-4k(-2k+5)\ge 0$$

$$k^2-4k+4+8k^2-20k \ge 0$$

$$9k^2-24k+4 \ge 0$$

$$ k \le \frac{24-\sqrt{24^2-4(4)(9)}}{18} \lor k \ge \frac{24+\sqrt{24^2-4(4)(9)}}{18}$$

$$ k \le \frac{4-\sqrt{4^2-4}}{3} \lor k \ge \frac{4+\sqrt{4^2-4}}{3}$$

$$ k \le \frac{4-2\sqrt{3}}{3} \lor k \ge \frac{4+2\sqrt{3}}{3}$$

Verification:

enter image description here

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  • $\begingroup$ Thank you so much $\endgroup$
    – Dima
    Sep 16, 2023 at 15:55

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