Find root of equation using bisection method?

Question

Question : Find an approximate value of $\sqrt[3]{25}$ using Bisection Method.

Since it doesnt state the accuracy in the question,how many iterations am I going to do to get that approximate value? Also since it doesnt state the interval,how am I even going to apply the bisection method on this one?

Answer 1

Let $f$ the function defined by: $\forall x\in\mathbb{R},\,f(x)=x^3-25$. We know that $f( \sqrt[3]{25})=0$ and the function has only one root since it's strictly increasing and continuous.

$f(0)$ for example is negative and $f(3)$ is positive so you can consider the interval $[0,3]$. You can take any other numbers.

The number of iterations is as you wish. As much it's bigger, the value you get is closer to $\sqrt[3]{25}$.

Answer 2

the accuracy is usually 3 decimal points or so.

you need to choose the function and the interval (they didn't give you function as well). the natural function to such a problem is $f(x)=x^3-25$, and since $2<25^{1/3}<3$, you can take the interval $[2,3]$