Option 4
f(x) = -x is an odd function
Solution:
A function is odd if and only if f(–x) = –f(x)
Option 1
[tex]f(x) = x^3 + 5x^2 + x[/tex]
Substitute x = -x in above equation
[tex]f(-x) = (-x)^3 + 5(-x)^2 + (-x)[/tex]
Cubes always involve multiplying a number by itself three times, so if the number is negative the cube will always be negative
Ans squaring results in positive
[tex]f(-x) = -x^3 + 5x -x[/tex] --- eqn 1
[tex]-f(x) = -(x^3 + 5x^2 + x)\\\\-f(x) = -x^3 - 5x^2 - x[/tex] ---- eqn 2
Comparing eqn 1 and eqn 2,
[tex]f(-x) \neq -f(x)[/tex]
Therefore not an odd function
Option 2
[tex]f(x) = \sqrt{x}[/tex]
[tex]f(-x) = \sqrt{-x}[/tex]
[tex]-f(x) = - \sqrt{x}[/tex]
Therefore,
[tex]f(-x) \neq -f(x)[/tex]
Therefore not an odd function
Option 3
[tex]f(x) = x^2 + x\\\\f(-x) = (-x)^2 + (-x)\\\\f(-x) = x^2 - x[/tex]
[tex]-f(x) = -(x^2 + x) = -x^2 - x[/tex]
[tex]f(-x) \neq -f(x)[/tex]
Therefore not an odd function
Option 4
[tex]f(x) = -x\\\\f(-x) = -(-x) = x[/tex]
[tex]-f(x) = -(-x) = x[/tex]
[tex]f(-x) = -f(x)[/tex]
Thus option 4 is correct and it is an odd function
