Help with Halving the interval/Bisection method! (1 Viewer)

[Sugar]

I'm a bit confused with determining where the root lies (hope that makes sense).

Say for example,
Show that x^3 + x^2 + x + 8 = 0 has a root between x = 1 and x = 2 and use the bisection method twice to determine the root.

Skipping the showing a root exists bit...

x = 1 and x = 2

xm = .5 (1 + 2) = 1.5
f(1.5) = -0.875 < 0

Since f(1.5) < 0, the root lies between 1.5 and 2. <-- I get confused as of this bit. Why does the root lie between 1.5 and 2 when f(1.5) < 0 ?

Continuing on,
f(1.75) = 2.18 > 0

Since f(1.75) > 0, the root lies between 1.5 and 1.75. <-- Same reason ^

Thanks in advance.

 

[velox]

Its because there is a sign change. I.e. the graph has gone from above the x axis to below the x axis hence the change of sign.

 

[who_loves_maths]

^ yes, continuing on from what velox said, the sign change implies the graph (which MUST be continuous) has gone from above/below the graph to below/above the graph - implying that it has cut the graph somewhere in between, hence a root exists in that local region.

however, remember that that graph must first be CONTINUOUS for 1.5 < x < 2 , since if it's not, then there's a possibility that there is no roots within that region even though there is a sign change... (eg. vertical asymptotes.)

 

[velox]

who_loves_maths added a good point about continuity there, however they dont usually have discontinous functions in these types of questions.

 

[who_loves_maths]

velox added a good point about the lax nature of HSC examinations there.

 

[velox]

They could be an ass, and make you do it on a piece meal function (whatever they are called), and make it a little more tricky that it is.

 

[who_loves_maths]

they don't need to, piecemeal will only make it too obvious. a normal discontinuous function (eg. rational functions) will do.