Question

Prove that the function is odd. \(f(x)=a_{2 n+1} x^{2 n+1}+\cdots+a_{3} x^{3}+a_{1} x\)

Short answer

Through substitution and simplification, it is proven that the given function is odd, since \(f(-x) = -f(x)\) for all values of \(x\) in the function's domain.

Step-by-step solution

Understanding the function

Observe the function \(f(x) = a_{2 n+1} x^{2 n+1}+\cdots+a_{3} x^{3}+a_{1} x\). Each term involves an odd power of \(x\) and a constant coefficient.

Substitution with -x

Substitute \(x\) with \(-x\) in the function: \(f(-x) = a_{2 n+1} (-x)^{2 n+1}+\cdots+a_{3} (-x)^{3}+a_{1} (-x)\). By this, every \(x\) in the function has been replaced with \(-x\).

Simplification

Simplify the expression. Each term of the function carries an odd exponent. As such, when \(-x\) is raised to these odd powers, it results in \(-x\). Therefore, the function simplifies to: \(f(-x)=-a_{2 n+1} x^{2 n+1}-\cdots -a_{3} x^{3} -a_{1} x\), which is \(-f(x)\). This proves that the function is odd.