Sketch products and quotients involving exponentials

The Teacher completes the blank Student asymptote working with $x\to-\infty$, $e^x\to0$ and $y\to0$, giving horizontal asymptote $y=0$. Independently, the inference needs the whole product: writing $t=-x\to\infty$ gives $e^x(x-1)=-(t+1)/e^t\to0$. It approaches from below. As $x\to+\infty$ the curve grows without bound. Its domain is all real $x$, with no vertical asymptote.

The source gives $x=0\Rightarrow y=-1$ and $y=0\Rightarrow x=1$, hence $(0,-1)$ and $(1,0)$. The Teacher supplies

$\frac{dy}{dx}=e^x(1)+(x-1)e^x=xe^x,\qquad xe^x=0\Rightarrow x=0.$

Independently $e^x>0$, so the derivative changes from negative to positive at $0$; the stationary point $(0,-1)$ is the minimum. The Teacher recommends demonstrating with TI-SmartView, showing differentiation if time permits, or using the graphing calculator to find the turn. In this lesson, predict the shape and use a calculator minimum search as a check, then compare with the exact derivative test.

The denominator $e^{2x}+1$ is positive for every real input. As $x\to-\infty$, $e^x,e^{2x}\to0$, so $y\to(-1)/1=-1$. Thus $y=-1$ is a horizontal asymptote on the left. The Teacher adds the curve and asymptote labels to the visible Student blank graph.

Both sources warn that the formal expression $\infty/\infty$ is not a limit value. Their complete manipulation divides numerator and denominator by $e^{2x}$:

$\begin{aligned}&y\\ & =\frac{(e^x-1)/e^{2x}}{(e^{2x}+1)/e^{2x}}\\ & =\frac{1/e^x-1/e^{2x}}{1+1/e^{2x}}\\ & =\frac{e^{-x}-e^{-2x}}{1+e^{-2x}}.\end{aligned}$

Now $x\to+\infty$ gives $y\to0/1=0$, so $y=0$ is a horizontal asymptote on the right. For the intercepts, $x=0$ gives $y=0$, while $y=0$ requires $e^x=1$, hence $x=0$. Both axes meet the curve at the origin.

The printed instruction is “Use G.C.”; the Teacher sketch labels a maximum $(0.881,0.207)$. Independently let $t=e^x>0$, so

$y=\frac{t-1}{t^2+1},\qquad\frac{dy}{dx}=\frac{t(-t^2+2t+1)}{(t^2+1)^2}.$

Only $t=1+\sqrt2$ is positive. The derivative changes $+\to-$ there, giving the exact maximum $\big(\ln(1+\sqrt2),(\sqrt2-1)/2\big)$. This agrees with the source’s rounded calculator labels. There is no minimum: $y+1=t(t+1)/(t^2+1)>0$, while the lower limit $-1$ is approached only as $x\to-\infty$.

The next source page gives this five-step procedure for the same quotient:

1. Enter $(e^x-1)/(e^{2x}+1)$ into Y₁, keeping both full numerator and denominator grouped.
2. Press 2nd → WINDOW to open TBLSET. Set TblStart to a large positive value, such as 100. ΔTbl is the increment (default 1); the screenshot uses automatic independent/dependent values.
3. Press 2nd → GRAPH for TABLE, then inspect increasing inputs. The source table for $x=100,\ldots,110$ displays tiny positive values (about $4\mathrm E{-44}$ down to $2\mathrm E{-48}$), suggesting $y=0$ at the right end.
4. Return with 2nd → WINDOW and set TblStart to a large negative value, such as −100.
5. Use 2nd → GRAPH and inspect decreasing inputs. The source display at approximately −103 through −93 rounds the outputs to −1, suggesting the left asymptote $y=-1$.

The Teacher describes this as a way to guess asymptotes. Finite tables and rounded entries are evidence for a conjecture, not proof that a value is reached or an end limit exists. Use the preceding algebraic limits to justify both asymptotes. A displayed −1 is rounded: independently $y+1=t(t+1)/(t^2+1)>0$ for $t=e^x>0$.