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Help, I don't understand the last one,
thank you.
Imgur: The most awesome images on the Internet
Help, I don't understand the last one,
thank you.
they want an interval of length $0.01$
i.e. $(a,b),~\ni b-a \leq 0.01$
such that there exists $x \in (a,b),~f(x) = \cos(x)-x^3 = 0$
One way of doing this is by "bracketing" the solution.
from your previous work you know that $x \in (0,1)$
take the midpoint of this interval, $\dfrac 1 2$
$f\left(\dfrac 1 2 \right) \approx 0.75 > 0$
so now we know our root is in the interval $\left(\dfrac 1 2, 1\right)$
take the midpoint of this interval, $~\dfrac 3 4$
$f\left(\dfrac 3 4 \right) \approx 0.31 > 0$
so the root must be in $\left(\dfrac 3 4 , 1\right)$
and again $f\left(\dfrac 7 8\right) \approx -0.03$
so the root must be in $\left(\dfrac 3 4 , \dfrac 7 8\right)$
and just keep narrowing down the interval in this fashion until the distance between the endpoints is less than or equal to $0.01$