Question

Prove that a polynomial function f of odd degree has at least one real root. Hint: It may help to consider first the case of a cubic, i.e., f(x)=a₀+a₁ x+a₂ x²+a₃ x³ where a₃ = 0.

Answer 1

To prove that a polynomial function of odd degree has at least one real root, we observe that as x approaches infinity or negative infinity, the sign of f(x) will change, ensuring a point at which f(x) equals zero, thereby confirming at least one real root by the Intermediate Value Theorem.

Step-by-step explanation:

To prove that a polynomial function f of odd degree has at least one real root, consider the behavior of the polynomial function as x approaches positive and negative infinity. For an odd-degree polynomial with a positive leading coefficient, as x approaches positive infinity, f(x) will also approach positive infinity, and as x approaches negative infinity, f(x) will approach negative infinity. Conversely, if the leading coefficient is negative, f(x) will approach negative infinity as x approaches positive infinity, and positive infinity as x approaches negative infinity.

Due to this behavior, by the Intermediate Value Theorem, which states that for any value between f(a) and f(b) where a and b are points in the domain of f and f is continuous, there must be some c between a and b such that f(c) is equal to that value. Considering that f(x) must cross the x-axis as it goes from positive to negative infinity (or vice versa), there must be at least one real root where f(x) = 0.