Question

Prove that the function is even. \(f(x)=a_{2 n} x^{2 n}+a_{2 n-2} x^{2 n-2}+\cdots+a_{2} x^{2}+a_{0}\)

Short answer

The function \(f(x)=a_{2 n} x^{2 n}+a_{2 n-2} x^{2 n-2}+\cdots+a_{2} x^{2}+a_{0}\) is even because \(f(x) = f(-x)\) for every value of \(x\).

Step-by-step solution

Write down the function

The function that needs to be proved even is given by: \(f(x)=a_{2 n} x^{2 n}+a_{2 n-2} x^{2 n-2}+\cdots+a_{2} x^{2}+a_{0}\).

Substitute x with -x and simplify

Replace each instance of x with -x to get \(f(-x)=a_{2 n} (-x)^{2 n}+a_{2 n-2} (-x)^{2 n-2}+\cdots+a_{2} (-x)^{2}+a_{0}\). Since the exponent of each x term is even, the result is positive, regardless of whether x is positive or negative. Therefore, \(f(-x)\) simplifies to the original function \(f(x)\).

Conclusion

Because we find that \(f(-x) = f(x)\), the function is even.