Answer:
[tex]x\in(-\infty,-2)\cup(5,\infty)[/tex]
Step-by-step explanation:
The duadratic function [tex]g(x)=x^2+20[/tex] begin to exceed the linear function [tex]f(x)=3x+30[/tex] when [tex]g(x)>f(x)[/tex]
Solve this inequality:
[tex]x^2+20>3x+30\\ \\x^2-3x+20-30>0\\ \\x^2-3x-10>0\\ \\x^2-5x+2x-10>0\\ \\x(x-5)+2(x-5)>0\\ \\(x-5)(x+2)>0[/tex]
This inequality is equivalent to
[tex]\left[\begin{array}{l}\left\{\begin{array}{l}x-5>0\\x+2>0\end{array}\right.\\ \\\left\{\begin{array}{l}x-5<0\\x+2<0\end{array}\right.\end{array}\right.\Rightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x>5\\x>-2\end{array}\right.\\ \\\left\{\begin{array}{l}x<5\\x<-2\end{array}\right.\end{array}\right.\Rightarrow \left[\begin{array}{l}x>5\\ \\x<-2\end{array}\right.[/tex]
Answer: [tex]x\in(-\infty,-2)\cup(5,\infty)[/tex]
