show that there is an odd degree polynomial of every odd degree with no critical points

Question

(in other words show that for any odd n there is a polynomial f(x) of degree n with no points x such that f(0)=0

Answer 1

Let C be a positive constant and let n be any integer (n=0, 1, 2, 3, 4, 5, ....), then f(x)=x²ⁿ⁺¹+Cx will be an odd polynomial of order 2n+1 with no critical points.

 

f'(x)=(2n+1)x²ⁿ+C

0=(2n+1)x²ⁿ+C

x²ⁿ=(-C)/(2n+1)

• x²ⁿ is a number raised to an even power. It must therefore be positive-valued for all x (or zero if x=0).
• Because C and n are non-negative, the value of (-C)/(2n+1) must be negative-valued for all combinations of C and n
• A non-negative expression can never equal a negative expression, so there is no solution to f'(x)=0 and thus f(x) does not have a critical point.