Decreasing curves and the largest tangent intercept · source reference

Printed question

The diagram shows a sketch of the curve $y=xe^{-x}$.

(i) By differentiation, find the range of values of $x$ for which the graph of $y=xe^{-x}$ is decreasing. [3]

(ii) Determine the range of values of $x$ for which the graph of $y=xe^{-x}$ is decreasing and concave downwards. [3]

(iii) The tangent to the curve at $P(x,y)$ meets the $y$-axis at $R(0,h)$. Express $h$ in terms of $x$, and find the greatest possible value of $h$. [6]

[2011/NYJC/Prelim/I/9]

Printed working

(i) $y=xe^{-x}$, $dy/dx=e^{-x}-xe^{-x}=e^{-x}(1-x)$. Decreasing ⇒ $dy/dx<0$ ⇒ $e^{-x}(1-x)<0$ ⇒ $1-x<0$ since $e^{-x}>0$ for all real $x$ ⇒ $x>1$.

(ii) $d^2y/dx^2=e^{-x}(-1)-e^{-x}(1-x)=e^{-x}(x-2)$. Concave downwards ⇒ $d^2y/dx^2<0$ ⇒ $e^{-x}(x-2)<0$ ⇒ $x<2$. Combining the number-line conditions gives $1<x<2$.

(iii) Gradient at $(x,y)$ is $e^{-x}(1-x)$. $R(0,h)$ is on the tangent at $P$:

$$e^{-x}(1-x)=\frac{h-xe^{-x}}{0-x}\Rightarrow h-xe^{-x}=-xe^{-x}(1-x)\Rightarrow h=x^2e^{-x}.$$ $$\frac{dh}{dx}=2xe^{-x}-x^2e^{-x}=xe^{-x}(2-x).$$

At maximum/minimum $dh/dx=0$, so $xe^{-x}(2-x)=0\Rightarrow x=2$ or $x=0$.

$x$	$2^-$	$2$	$2^+$
$dh/dx$	$+$	$0$	$-$
Tangent-height trend	rising	horizontal	falling

$x$	$0^-$	$0$	$0^+$
$dh/dx$	$-$	$0$	$+$
Tangent-height trend	falling	horizontal	rising

Printed conclusion: maximum $h$ occurs at $x=2$, greatest possible $h=4e^{-2}$.

Source-domain qualification

No $x\ge0$ restriction is printed, and the source sketch includes negative $x$. On all real $x$, $x^2e^{-x}\to\infty$ as $x\to-\infty$, so there is no greatest value. $4e^{-2}$ is a local maximum and is the greatest value on $x\ge0$. At $x=0$ the intercept formula follows directly from the tangent, though the displayed secant-gradient rearrangement divides by $x$.

Answer key

(i) $x>1$; (ii) $1<x<2$; (iii) printed $4e^{-2}$, qualified above.

Native diagrams