Answer:
The statement that is true about functions is:
If f and g are function then,
B. (f+g)(x)=f(x)+g(x)
Step-by-step explanation:
A)
Let us define f(x)=x² and g(x)=1
then we have:
[tex]\dfrac{f(x)}{g(x)}=x^2[/tex]
and
[tex]\dfrac{g(x)}{f(x)}=\dfrac{1}{x^2}[/tex]
Hence, we get:
[tex]\dfrac{f(x)}{g(x)}\neq \dfrac{g(x)}{f(x)}[/tex]
Hence, option: A is incorrect.
B)
Option: B is always true.
(f+g)(x)=f(x)+g(x)
C)
Let us define f(x)=x²
Then
[tex]f(a+b)=(a+b)^2=a^2+b^2+2ab[/tex]
and
[tex]f(a)+f(b)=a^2+b^2[/tex]
Hence, we get:
[tex]f(a+b)[/tex] is not always equal to [tex]f(a)+f(b)[/tex]
D)
Let us suppose [tex]f(x)=4x\ and\ g(x)=\dfrac{1}{x}[/tex]
Now,
[tex](fog)(1)=f(g(1))\\\\\\(fog)(1)=f(1)\\\\\\(fog)(1)=4[/tex]
and
[tex](gof)(1)=g(f(1))\\\\\\(gof)(1)=g(4)\\\\\\(gof)(1)=\dfrac{1}{4}[/tex]
Hence, we get:
[tex](fog)(x)[/tex] is not always equal to [tex](gof)(x)[/tex]
