Bisection steps

Let f(x) = x^3 - 2x - 1; a graph of f is shown below. Suppose we apply three bisection steps to f over the interval [0,2]. We will then decide that the root of f lies in what interval?

Every time we apply a bisection step, the interval halves. Notice that the length of our initial interval is 2-0=2.
Thus our first would have length \frac{1}{2} (2) =1.
Second would have \frac{1}{2} (1)=\frac{1}{2}
Third would have length \frac{1}{2}\big(\frac{1}{2}\big)=\frac{1}{4}.

Since the root appears to be closer to 2 than to 0, the first step of the bisection method will result in the interval [1,2].

Since the root appears to be closer to 2 than to 1, the next step of the bisection method will produce the interval [1.5,2]

Since the root appears to be closer to 1.5 than to 2, the third step of the bisection method will result in the interval [1.5,1.75].