Show that the function f defined by
F(x)=x^3+4x^2+x-6
has a real root in the interval [0,2]
Solution.
Domain(f)=(−∞,∞).(-\infty,\infty).(−∞,∞).
f(0)=−6<0.f(0)=-6<0.f(0)=−6<0.
f(2)=8+16+2−6=20>0.f(2)=8+16+2-6=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).x=1,1∈(0,2).
f(1)=1+4+1−6=0.f(1)=1+4+1-6=0.f(1)=1+4+1−6=0.
So, x=1x=1x=1 is real root of function f(x) on interval (0,2).
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