I am not understanding what has asked to compute of the following exercise.

Question

let $f(x)=(x+2)(x+1)x(x-1)^3(x-2)$.

To which zero of $f$ does the Bisection method converges when applied on the interval $[-3,2.5]$

Have i asked to find the root of $f(x)$ ?

Answer 1

Do you know how to implement the bisection method? Every zero of $f$ lies in the initial interval (how can we tell?), but after a few iterations, there will be only one remaining. That is the zero to which the method will converge.

Answer 2

Yes, you are right. Bisection method is a method of finding roots of a function.

Answer 3

To avoid evaluating the actual function repeatedly, note first that the zeros of the function are $$x=-2, -1, 0, 1, 2$$ ($x=1$ is a triple zero)

These are all "trapped" by the original boundaries. Note also that the function changes sign at each zero; the exponent of $3$ on the $(x-1)$ term ensures that. Also, with an odd number of zeros, the function has a different sign at each original boundary.

The first trial is at $$x_1=\frac{-3+2.5}{2}=-0.25$$This "traps" the zeros $-1$ and $-2$ to the left in the interval $(-3, -0.25)$; the sign of the function is the same at both ends of this sub-interval.

$x_1=-0.25$ traps the other three zeros to the right in the interval $(-0.25, 2.5)$; the function has a different sign at $-0.25$ and $2.5$. This means that the next trial is at:$$x_2=\frac{-0.25+2.5}{2}=1.125$$This traps two zeros between $-0.25$ and $1.125$; the function has the same sign at each end of this sub-interval.

$x_2=1.125$ traps the one remaining zero, $2$, in the interval $(1.125, 2.5)$. This is the zero that will ultimately be determined from the original interval.