Calculus I · Limits and Continuity · lesson

Bisection Method After the IVT

Visual study stop

Read the picture before the symbols

Pause at each graph long enough to say what the input is doing, what the output is doing, and which feature supports the next mathematical decision.

Continuous cubic crossing the x-axis between endpoints of opposite sign. — **Keep the guaranteed crossing inside the current bracket**
The endpoint signs guarantee at least one crossing because the function is continuous. Each bisection step tests a midpoint and preserves the half-interval whose endpoint signs still differ.

Concept

Learning objectives

Use repeated sign changes on smaller intervals to approximate a root guaranteed by the Intermediate Value Theorem.

Bisection: Turning Existence Into an Approximation

For

f(x)=x^3+x-1,

we know a root lies in $(0,1)$ . Test the midpoint $0.5$ :

f(0.5)=0.125+0.5-1=-0.375.

Since $f(0.5)<0$ and $f(1)>0$ , a root lies in $(0.5,1)$ .

Test the new midpoint $0.75$ :

f(0.75)=0.421875+0.75-1=0.171875>0.

Now the root lies in $(0.5,0.75)$ .

Continue:

Reference table
$\text{interval}$	$\text{midpoint}$	$\text{sign at midpoint}$
$(0,1)$	$0.5$	$-$
$(0.5,1)$	$0.75$	$+$
$(0.5,0.75)$	$0.625$	$-$
$(0.625,0.75)$	$0.6875$	$+$
$(0.625,0.6875)$	$0.65625$	$-$

The root is being trapped in shorter intervals. This is the bisection method.

Concept

The Intermediate Value Theorem says a root is in the box. Bisection keeps cutting the box in half and throwing away the half that cannot contain the sign change.

After the explanation

Use the section idea

Reading lens

Do the limit, the function value, and the surrounding domain fit together at the point or across the interval?

Mental model

Continuity is a three-part agreement: the value exists, the two-sided limit exists, and those two quantities are equal.

Decision

At a point, test the three conditions in order; on an interval, check the domain and endpoints before invoking any continuity theorem.

Common trap

A sign change supports the Intermediate Value Theorem only when continuity holds on the entire closed interval, and it does not prove uniqueness.

Check yourself

You are ready to continue when you can classify a break, decide whether one value can repair it, and state every IVT hypothesis aloud.

Source & rights

Original instruction with traceable references.

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