1. ## continuity problem

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Help, I don't understand the last one,

thank you.

2. ## Re: continuity problem

Originally Posted by ihatecalculus
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Help, I don't understand the last one,
Why are you so lazy as to not just type out the actual question.
Your link contains awful SPAM. How dare you put us through that?

3. ## Re: continuity problem

Originally Posted by ihatecalculus
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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 $0.01$