Question 734675
a) Consider {{{f(x)=cos(x)-x^3}}}
It is a continuous function.
{{{f(0)=cos(0)-0^3=1>0}}}
{{{f(pi/2)=cos(pi/2)-(pi/2)^3=0-pi^3/8=-pi^3/8<0}}}
The continuous function {{{f(x)}}} is positive for {{{x=0}}} and negative for {{{x=pi/2}}} so it must have a zero somewhere in the (0,{{{pi/2}}}) interval (according to the intermediate value theorem).
 
You could do something similar with {{{g(x)=x^5-x^2+2x+3>0}}}.
You can easily see that {{{g(0)=3>0}}} and {{{g(-10)=-100000+100+20+3<0}}} and that is as much as is needed to prove is that there is one real root.