Given:
[tex]f(x)=x^{3}+5 x^{2}+x[/tex]
[tex]f(x)=\sqrt{x}[/tex]
[tex]f(x)=x^{2}+x[/tex]
[tex]f(x)=-x[/tex]
To find:
The odd function.
Solution:
If f(-x) = -f(x), then the function is odd.
If f(-x) = f(x), then the function is even.
Option A: [tex]f(x)=x^{3}+5 x^{2}+x[/tex]
Substitute x = -x
[tex]f(-x)=(-x)^{3}+5 (-x)^{2}+(-x)[/tex]
[tex]=-x^3+5x^2-x[/tex]
[tex]=-(x^3-5x^2+x)[/tex]
≠ - f(x)
It is not odd function.
Option B: [tex]f(x)=\sqrt{x}[/tex]
Substitute x = -x
[tex]f(-x)=\sqrt{-x}[/tex]
[tex]=\sqrt{x} i[/tex] (Since [tex]\sqrt{-1} =i[/tex])
≠ - f(x)
It is not odd function.
Option C: [tex]f(x)=x^{2}+x[/tex]
Substitute x = -x
[tex]f(-x)=(-x)^{2}+(-x)[/tex]
[tex]=x^2-x[/tex]
[tex]=-(-x^2+x)[/tex]
≠ - f(x)
It is not odd function.
Option D: [tex]f(x)=-x[/tex]
Substitute x = -x
[tex]f(-x)=-(-x)[/tex]
[tex]=-f(x)[/tex]
It is odd function.
Therefore, f(x) = -x is an odd function.
