It's up to you to find to values of $x$ such that the function value changes sign.
It will be easy trial and error. (unlikely in an actual test)
Let's randomly (with a bit of intuition) try $x=0$ and $x=2$!
$f(x)=x^3+3x-7$
$f(0)=-7<0$
$f(2)=2^3+3(2)-7=8+6-7=7>0$
Since $f(0)<0$ and $f(2)>0$, then there is at least one root between $x=0$ and $x=2$.
Now proceed to half the interval 3 times.....
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