When is the tangent line of the following function horizontal?
$$y =\frac{\sin x}{e^x}$$
What steps or how can I draw a graph to figure this out or prove this?
$\endgroup$ 12 Answers
$\begingroup$Apply the quotient rule to find $$ y' = \frac{e^x \cos x - e^x \sin x}{e^{2x}} $$
We want $y'=0$, which means we want the numerator to equal zero:
$$ 0 = e^x \cos x - e^x \sin x $$ $$ 0 = \cos x - \sin x $$ $$ \sin x = \cos x $$ Divide by $\cos x$: $$ \tan x = 1 $$
This happens at $x = \frac{\pi}{4}$, and since tangent has period $\pi$, it will happen also when we add an integer multiple of $\pi$. Thus, we have a horizontal tangent line at $$ x = \frac{\pi}{4} + n\pi, n \in \mathbb{Z} $$
$\endgroup$ 6 $\begingroup$If we have a function $y = f(x)$, then the derivative of $y$ with respect to $x$ at the point $x_0$ gives us the slope of the tangent at $x_0$. If you want the tangent to be horizontal, this means that the slope of the tangent must be $0$. Hence, we want to find $x_0$ such that $\left. \dfrac{dy}{dx} \right \rvert_{x_0} = 0$.
In your case, we have $y = \sin(x) \exp(-x)$. Hence, for the tangent to be horizontal at $x_0$, we need $$\left. \dfrac{dy}{dx} \right \rvert_{x_0} = -\sin(x_0) \exp(-x_0) + \cos(x_0) \exp(-x_0) = 0 $$ Note that $\exp(-x) >0, \forall x \in \mathbb{R}$. Hence, we get that $$- \sin(x_0) + \cos(x_0) = 0 \implies \tan(x_0) = 1 \implies x_0 = n\pi + \dfrac{\pi}4, \, \forall n \in \mathbb{Z}$$ Hence, at all these points i.e. at $x_0 = n\pi + \dfrac{\pi}4, \, \forall n \in \mathbb{Z}$, we have the tangent being horizontal.
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