Answer to Question #190650 in Real Analysis for Nikhil

Question #190650

Show that the function f defined by

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

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


1
Expert's answer
2021-05-10T18:23:01-0400

Solution.


f(x)=x3+4x2+x6f(x)=x^3+4x^2+x-6

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

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

f(2)=8+16+26=20>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(0,2).x=1, 1 \in (0,2).

f(1)=1+4+16=0.f(1)=1+4+1-6=0.

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


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