I need to find the range of a function:

$$f(x) = \frac{e^{2x} - e^x + 1}{e^{2x} + 3e^x - 7}$$

Substituting $$e^x = t$$ and equating $$f(x)$$ to $$y$$, I get

$$(y-1)t^2 + (3y+1)t - (7y+1) = 0$$

Since $$t$$ must be real and positive, I get the range of y to be $$(-\infty, -\frac{1}{3})\bigcup(1, \infty)$$.

Is this correct?

• Your reasoning is correct. The calculations are $\dfrac{-b}{2a}>0$ and $b^2-4ac>0$. I didn’t check your final answer. Commented Sep 6, 2022 at 16:07
• The final range is wrong. Commented Sep 6, 2022 at 17:06
• For $x \to - \infty$, the expression approaches $-1/7$, so your range isn't quite right. The reasoning appears to be solid. Commented Sep 6, 2022 at 19:58

Set $$f(t) = \frac{t^2-t+1}{t^2+3t-7}$$. We want the range of $$f$$ on $$(0,\infty)$$. This function has a singularity (vertical asymptote) at $$t_0 = \frac{\sqrt{37} - 3}{2}$$. Now $$f'(t) = \frac{4 \left(t^2-4 t+1\right)}{\left(t^2+3 t-7\right)^2}.$$ Then $$f'(t) = 0$$ for $$t_{1,2} = 2 \mp \sqrt{3}$$. and therefore $$f' > 0$$ on $$(0,t_1)$$, $$f' < 0$$ on $$(t_1,t_0) \cup (t_0,t_2)$$ and $$f' > 0$$ on $$(t_2,\infty)$$.
It follows that $$f((0,\infty)) = (-\infty, f(t_1)] \cup [f(t_2,\infty)$$ which is the range given in the previous answer.
Substitute $$f(x)$$ with $$y$$: $$y=\frac{e^{2x}-e^x+1}{e^{2x}+3e^x-7}$$. Rewrite in terms of $$e^x$$:
$$e^{2x}y+3e^xy-7y=e^{2x}-e^x+1$$ $$(y-1)(e^x)^2+(3y+1)e^x-(7y+1)=0$$ This equation has roots (with respect to $$x$$), when there are roots with respect to $$e^x$$ (discriminant is non-negative) and at least one root is positive.
$$D=(3y+1)^2-4(y-1)(-7y-1)=37y^2-18y-3\ge0$$ or $$y\in\left(-\infty, \frac{9-8\sqrt{3}}{37}\right]\cup\left[\frac{9+8\sqrt{3}}{37}, \infty\right)$$
In this case $$e^x=\frac{-(3y-1)\pm\sqrt{37y^2-18y-3}}{2(y-1)}$$