Respuesta :
If the [tex]f(x)[/tex] and [tex]t(x)[/tex] are even function then [tex]fo\ t\ (x)[/tex] is an even function, if [tex]f(x)[/tex] and [tex]t(x)[/tex] are odd function then the function [tex]fo\ t\ (x)[/tex] is an odd function and if [tex]f(x)[/tex] is even and [tex]t(x)[/tex] is odd then the function [tex]fo\ t\ (x)[/tex] is an even function.
Further explanation:
An even functrion satisfies the property as shown below:
[tex]\boxed{f(-x)=f(x)}[/tex]
An odd functrion satisfies the property as shown below:
[tex]\boxed{f(-x)=-f(x)}[/tex]
Consider the given composite function as follows:
[tex]\boxed{fo\ t\ (x)=f\left(t(x))\right}[/tex]
If both the function [tex]f(x)[/tex] and [tex]t(x)[/tex] are even function.
[tex]\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\\&=f\left(t(x))\right\\&=fo\ t\ (x)\end{aligned}[/tex]
From the above calculation it is concluded that,
[tex]\boxed{fo\ t\ (-x)=fo\ t\ (x)}[/tex]
This implies that the composite function [tex]fo\ t\ (x)[/tex] is an even function.
If both the function [tex]f(x)[/tex] and [tex]t(x)[/tex] are odd function.
[tex]\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\\&=f\left(-t(x))\right\\&=-fo\ t\ (x)\end{aligned}[/tex]
From the above calculation it is concluded that,
[tex]\boxed{fo\ t\ (-x)=-fo\ t\ (x)}[/tex]
This implies that the composite function [tex]fo\ t\ (x)[/tex] is an odd function.
If the function [tex]f(x)[/tex] is even and [tex]t(x)[/tex] is odd.
[tex]\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\\&=f\left(-t(x))\right\\&=fo\ t\ (x)\end{aligned}[/tex]
From the above calculation it is concluded that,
[tex]\boxed{fo\ t\ (-x)=fo\ t\ (x)}[/tex]
This implies that the composite function [tex]fo\ t\ (x)[/tex] is an even function.
