Answer:
The function touches the damping factor
at x=[tex]\frac{(4n-3)\pi}{2}[/tex] and x=[tex]\frac{(4n-1)\pi}{2}[/tex]
The x-intercept of f(x) is
at x=[tex]n\pi[/tex]
Step-by-step explanation:
Given function is f(x)=[tex]e^{-3x} sin(x)[/tex] and damping factor as y=[tex]e^{-3x}[/tex] and y=[tex](-1)e^{-3x}[/tex]
To find when function touches the damping factor:
For f(x)=[tex]e^{-3x} sin(x)[/tex] and y=[tex]e^{-3x}[/tex]
Equating the both the equation,
[tex]e^{-3x} sin(x)=e^{-3x}[/tex]
[tex] sin(x)=1[/tex]
x=[tex]\frac{(4n-3)\pi}{2}[/tex]
For f(x)=[tex]e^{-3x} sin(x)[/tex] and y=[tex](-1)e^{-3x}[/tex]
Equating the both the equation,
[tex]e^{-3x} sin(x)=(-1)e^{-3x}[/tex]
[tex] sin(x)=(-1)[/tex]
x=[tex]\frac{(4n-1)\pi}{2}[/tex]
Therefore, The function touches the damping factor x=[tex]\frac{(4n-3)\pi}{2}[/tex] and x=[tex]\frac{(4n-1)\pi}{2}[/tex]
To find x-intercept of f(x):
For x-intercept, y=0
f(x)=[tex]e^{-3x} sin(x)[/tex]
y=[tex]e^{-3x} sin(x)[/tex]
[tex]e^{-3x} sin(x)=0[/tex]
Hence, [tex]e^{-3x}[/tex] is always greater than zero.
Therefore,[tex]sin(x)=0[/tex]
x=[tex]n\pi[/tex]
Thus,
The x-intercept of f(x) is at x=[tex]n\pi[/tex]
