Answer to Question #190650 in Real Analysis for Nikhil

Question

Show that the function f defined by

F(x)=x^3+4x^2+x-6

has a real root in the interval [0,2]

Answer 1

Solution.

$$f(x)=x^3+4x^2+x-6$$

Domain(f)=$(-\infty,\infty).$

$f(0)=-6<0.$

$f(2)=8+16+2-6=20>0.$

There are must be at least one real root between 0 and 2.

Check point $x=1, 1 \in (0,2).$

$f(1)=1+4+1-6=0.$

So, $x=1$ is real root of function f(x) on interval (0,2).