Question

Show that the curves \(y = {e^{ - x}}\) and \(y = - {e^{ - x}}\) touch the curve \(y = {e^{ - x}}\sin {\kern 1pt} x\) at its inflection points.

Short answer

It is proved that the curves \(y = {e^{ - x}}\) and \(y = - {e^{ - x}}\) touch the curve \(y = {e^{ - x}}\sin x\) at its inflection points.

Step-by-step solution

Given data

The given function is \(y = {e^{ - x}}\sin {\kern 1pt} x\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Simplify the expression

The expression is: \(y = {e^{ - x}}\sin x\) meets \(y = {e^{ - x}}\) when \({e^{ - x}}\sin x = {e^{ - x}}\)

We can divide both sides by \({e^{ - x}}\) because it is never 0; \(\sin x = 1\)

This implies \(x = \frac{\pi }{2} + 2n\pi \) ………. (1)

Two curves touch each other if and only if their slopes are equal

Therefore, we need to ensure that the derivatives of \({e^{ - x}}\sin x\) and \({e^{ - x}}\) are equal

\({e^{ - x}}(\cos x - \sin x) = - {e^{ - x}}\)

Divide both sides by \({e^{ - x}}\): \(\cos x - \sin x = - 1\)

The above equation is true when \(x = \frac{\pi }{2} + 2n\pi \) because it makes \(\cos x = 0\) and \(\sin x = 1\).

Divide both the sides

The expression is: \(y = {e^{ - x}}\sin x\) meets \(y = - {e^{ - x}}\) when \({e^{ - x}}\sin x = - {e^{ - x}}\)

We can divide both sides by \({e^{ - x}}\) because it is never 0, \(\sin x = - 1\)

This implies \(x = \frac{\pi }{2} + 2n\pi \) ……… (2)

Two curves touch each other if and only if their slopes are equal

Therefore, we need to ensure that the derivatives of \({e^{ - x}}\sin x\) and \( - {e^{ - x}}\) are equal

\({e^{ - x}}\left( {\cos x - \sin x} \right) = {e^{ - x}}\)

Divide both sides by \({e^{ - x}}\): \(\cos x - \sin x = 1\)

The above equation is true when \(x = - \frac{\pi }{2} + 2n\pi \) because it makes

\(\cos x = 0\) and \(\sin x = - 1\).

Differentiate the expression

We have now shown that \(y = \pm {e^{ - x}}\) touch \(y = {e^{ - x}}\sin x\) at \(x = \pm \frac{\pi }{2} + 2n\pi \)

The only thing left is to show that these points are inflection points for \(y = {e^{ - x}}\sin x\)

We have already found that: \({y^{\prime \prime }} = {e^{ - x}}(\cos x - \sin x)\)

Differentiate using the product rule: \({y^{\prime \prime }} = - {e^{ - x}}\left( {\cos x - \sin x} \right) + {e^{ - x}}\left( { - \sin x - \cos x} \right)\)

Take out \({e^{ - x}}\) as the common factor

\(\begin{array}{c}{y^{\prime \prime }} = {e^{ - x}}\left( { - \cos x + \sin x - \sin x - \cos x} \right)\\{y^{\prime \prime }} = - 2{e^{ - x}}\cos x\end{array}\)

Since \(\cos x\) changes its sign when \(x = \pm \frac{\pi }{2} + 2n\pi \), we can conclude that these points are also inflection points.

It is proved that the curves \(y = {e^{ - x}}\) and \(y = - {e^{ - x}}\) touch the curve \(y = {e^{ - x}}\sin x\) at its inflection points.