Question

Examine whether the equation, x^3-11x+9=0 has a real root in the interval, [-1,2]

Answer 1

Solution:

$x^3-11x+9=0$

$f(x)=x^3-11x+9$

Interval: [-1,2]

$f(-1)=(-1)^3-11(-1)+9 \\=-1+11+9 \\=19 \\f(2)=(2)^3-11(2)+9 \\=8-22+9 \\=-5$

Now, $f(-1).f(2)=19(-5)=-95<0$

$\because f(-1).f(2)<0$

So, yes, a real root exists in the interval [-1,2]