Complete study of sin x · e^{-cos x}

Problem

Carry out the complete study of the function $f(x) = \sin x\cdot e^{-\cos x}.$

Solution

Step 1 — Domain. $f$ is defined for every $x\in\mathbb{R}$ (a product of functions defined everywhere). $f$ is periodic with period $2\pi$.

Step 2 — Symmetries. $f(-x) = \sin(-x)\cdot e^{-\cos(-x)} = -\sin x\cdot e^{-\cos x} = -f(x).$ Hence $f$ is odd (graph symmetric with respect to the origin). It is enough to study $f$ on $[0,2\pi]$ (or on $[0,\pi]$ and reconstruct by symmetry).

Step 3 — Sign. $e^{-\cos x}>0$ always (it is an exponential). So $f(x)>0 \iff \sin x > 0 \iff x\in(0,\pi)$ (in the first period). $f(x)<0$ for $x\in(\pi,2\pi)$, while $f(x)=0$ for $x=0, \pi, 2\pi$.

Sign of $f$ in the first period: positive on $(0,\pi)$, negative on $(\pi,2\pi)$.

Step 4 — First derivative. With the product rule and the derivative of the composite: $f'(x) = \cos x\cdot e^{-\cos x} + \sin x\cdot e^{-\cos x}\cdot\sin x = e^{-\cos x}(\cos x + \sin^2 x).$ Since $e^{-\cos x}>0$, the sign of $f'$ is that of $\cos x + \sin^2 x = \cos x + 1 - \cos^2 x$. Setting $u=\cos x$ we must study $-u^2+u+1 > 0 \iff u^2-u-1<0, \qquad u\in[-1,1].$ With $\Delta = 1+4 = 5$ we obtain $u = \dfrac{1\pm\sqrt{5}}{2}$. The only root in $[-1,1]$ is $u_0 = \frac{1-\sqrt{5}}{2}\approx -0{,}618.$ Hence $f'(x)>0$ when $\cos x > u_0$, that is when $x$ is “close enough” to $0$ or to $2\pi$; $f'(x)<0$ otherwise. In $(0,2\pi)$ we have $\cos x = u_0$ for $x\approx 2{,}24$ and $x\approx 4{,}04$. The stationary points are therefore $x\approx 2{,}24 \ (\text{relative maximum}), \qquad x\approx 4{,}04 \ (\text{relative minimum}).$

Step 5 — Second derivative. Left as an exercise; the sign of $f''$ determines the inflection points.

Step 6 — Graph.

Graph of $f(x)=\sin x\cdot e^{-\cos x}$ on $[0,2\pi]$: relative maximum $\approx 1{,}46$ near $x\approx 2{,}24$, relative minimum $\approx -1{,}46$ near $x\approx 4{,}04$. The odd function extends by symmetry with respect to the origin.

Links

Topics: Curve sketching
Concepts: First derivative · Domain · Relative maxima and minima · Sign · Symmetries · Curve sketching
Methods: Rules of differentiation · Maximum and minimum by derivative · Curve sketching
Skills: Differentiating · Sketching a function · Drawing a graph
Exercise type: Curve sketching