Prove function has exactly one real root

Example: Prove $f(x)=3x^5+\pi x^3+7x-4$

Proof:

1. Show there exists a real root
   1. $f(0)=-4$
   2. $f(1)=\pi +6$
   3. note that $-4<0<\pi +6$
   4. Then, there exists a $c\in (0,1)$ such that $f(c)=0$ by IVT
2. Show that f has at most one real root
   1. Suppose f has two real roots. Assume $f(a)=f(b)=0$
   2. Note that $f$ is cont on $R$ as $f$ is a polynomial
      1. Thus $f$ is cont on $[a,b]$ as $[a,b]\subset R$
   3. Note that $f$ is diff on $R$ as $f$ is a polynomial
      1. Hence $f$ is diff on $[a,b]$ as $[a,b]\subset R$
   4. Note that $f(a)=f(b)$ by assumption
   5. By Rolle’s Theorem, there is a $c\in (a,b)$ such that $f^{\prime }(c)=0$
   6. Note that $f^{\prime }(x)=15x^4+3\pi x^2+7>0+0+7>0$ for all $x\in R$
   7. 5 and 6 are a contradiction
   8. Therefore $f$ has at most one real roots $\square$