if $kgeq1$, the graphs of $y = sin x$ and $y = ke^{-x}$ intersect for $xgeq0$. find the smallest value of…

Explanation:

Step1: Differentiate the functions

The derivative of $y = \sin x$ is $y'=\cos x$, and the derivative of $y = ke^{-x}$ is $y'=-ke^{-x}$.

Step2: Set up equations for tangent - point

Since the two graphs are tangent, at the point of tangency $x = a$, we have $\sin a=ke^{-a}$ (equating the $y$ - values) and $\cos a=-ke^{-a}$ (equating the slopes).

Step3: Eliminate $k$ and $e^{-a}$

From $\sin a=ke^{-a}$ and $\cos a=-ke^{-a}$, we get $\sin a=-\cos a$. So, $\tan a=- 1$. For $x\geq0$, the smallest non - negative value of $a$ for which $\tan a=-1$ is $a=\frac{3\pi}{4}$.

Step4: Find the value of $k$

Substitute $x = \frac{3\pi}{4}$ into $\sin x=ke^{-x}$. We have $\sin\frac{3\pi}{4}=ke^{-\frac{3\pi}{4}}$. Since $\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$, then $\frac{\sqrt{2}}{2}=ke^{-\frac{3\pi}{4}}$, and $k=\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}$.

Step5: Find the $y$ - coordinate of the point of tangency

Substitute $x = \frac{3\pi}{4}$ into $y = \sin x$. So, $y=\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$.

Answer:

$k=\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}$ $x=\frac{3\pi}{4}$ $y=\frac{\sqrt{2}}{2}$