From the graph of $f$ in Figure 1.7.1(a), it would appear that there are three zeros located one each in the following three intervals: $\left[-1\,-0.5\right]\,\left[0.5\,0.9\right]\,\left[1.1\,1.5\right]$.

Note how the vertical asymptote at $x\=1$ is avoided as an endpoint of an interval that "traps" a zero.

The three zeros are, in fact, $-0.6807685310$, $0.7471268154$, and $1.450025320$, as found by Maple's floating-point solver. These values certainly fall within the three intervals obtained from Figure 1.7.1(a).

Below, the bisection algorithm is detailed, and its connection to the Intermediate Value theorem explained.$$

Control-drag (or type) $f\left(x\right)\=\dots$ 
Context Panel: Assign Function

Leftmost interval

Middle interval

Rightmost interval

If the product $f\left(a\right)\cdot f\left(b\right)<0$, then $f\left(a\right)$ and $f\left(b\right)$ differ in sign.
If the product is zero, then one of $a$ or $b$ is a zero!

If the sign of a continuous function changes at the endpoints of an interval $\left[a\&comma;b\right]$, then, by the Intermediate Value theorem, the function must take the value zero somewhere between $a$ and $b$. This is the essence of the idea behind the bisection method.

In Maple's implementation of the method, the function is evaluated at the midpoint $p\&equals;\left(a\&plus;b\right)\&sol;2$, and the half-interval in across which there continues to be a sign change is retained. This process is repeated until the zero is "trapped" in a sufficiently small interval. The midpoint of this last interval is then taken as the approximation to the zero.

Tables 1.7.1(a-c) display the output of Maple's Bisection command used to approximate each of the three zeros of the given function $f$.

interface(rtablesize=100):

Student:-NumericalAnalysis:-Bisection(f(x),[-1,-.5],output=information,maxiterations=50,stoppingcriterion=function_value,tolerance=1e-6);

interface(rtablesize=100):

Student:-NumericalAnalysis:-Bisection(f(x),[0.5,0.9],output=information,maxiterations=50,stoppingcriterion=function_value,tolerance=1e-7);

interface(rtablesize=100):

Student:-NumericalAnalysis:-Bisection(f(x),[1.1,1.5],output=information,maxiterations=50,stoppingcriterion=function_value,tolerance=1e-7);