Which of the following is an odd function?
To determine which function is an odd function, we recall the definition: a function \(f(x)\) is odd if and only if \(f(-x) = -f(x)\) for all \(x\) in its domain. Conversely, a function is even if \(f(-x) = f(x)\). We will test each of the four given options to see which one satisfies the definition of an odd function.
We test each function:
\(f(-x) = (-x)^2 + (-x) = x^2 - x\).
\(-f(x) = -(x^2 + x) = -x^2 - x\).
Since \(f(-x)
eq -f(x)\), this function is not odd.
\(f(-x) = (-x)^3 + 5(-x)^2 + (-x) = -x^3 + 5x^2 - x\).
\(-f(x) = -(x^3 + 5x^2 + x) = -x^3 - 5x^2 - x\).
Since \(f(-x)
eq -f(x)\), this function is not odd.
The domain is \(x \geq 0\). For negative \(x\), \(-x\) is not in the domain (assuming real numbers). Thus, this function does not satisfy the symmetry condition required to be odd.
\(f(-x) = -(-x) = x\).
\(-f(x) = -(-x) = x\).
Since \(f(-x) = -f(x)\), this function is odd.
After evaluating each function against the definition of an odd function, we found that only the function \(f(x) = -x\) satisfies \(f(-x) = -f(x)\). Thus, option (d) is the correct answer.
Odd functions are frequently used in physics and signal processing to represent phenomena with rotational symmetry. For instance, in an alternating current (AC) circuit, the current can be modeled as an odd function because the current flow is symmetric about the origin of the time axis. Understanding whether a function is odd or even helps engineers simplify calculations when integrating signals over symmetric intervals, as the integral of an odd function over a symmetric interval \([-a, a]\) is always zero, which simplifies the analysis of wave interference and circuit efficiency.