Question: Show that every polynomial of odd degree with real coefficients has at least one real root.
Proof: f(x) is of the form:
f(x)=anxn+an−1xn−1+.....+a2x2+a1x+a0
f(x) is an odd degree polynomial and is hence continuous on R.
Then, if f(x) has a positive leading coefficient
⟹lim and \lim_{x \to -\infty}f(x)=-\infty
\implies \exists \alpha,\beta \in\mathbb{R} with \alpha > \beta such that:
\alpha>0 and f(\alpha)>0 and \beta <0 and f(\beta)<0
Now, consider the restriction of f(x) onto the interval I:=[\beta,\alpha]
The restriction of f on I is also continuous on I
Then, since $f(\beta)<0
\exists a number c \in (\beta,\alpha) such that f(c) =0
Can anyone please verify this proof and let me know if it is correct and /or if I'm missing out on something?
Note: I've missed out on the case with negative leading coefficient as I believe that too will be done in a similar fashion should this proof be alright.
Thank you.
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