The Intermediate Value Theorem
The Intermediate Value Theorem:
Suppose 𝑓 is a continuous function on the closed interval 𝑎 ≤ 𝑥 ≤ 𝑏 (in other words: [𝑎, 𝑏]),
and suppose 𝑁 is some number that is between the two endpoints 𝑓(𝑎) and 𝑓(𝑏), where
𝑓(𝑎) ≠ 𝑓(𝑏). Then there exists a number 𝑐 such that 𝑎 < 𝑐 < 𝑏 and 𝑓(𝑐) = 𝑁.
Interpretation: you’ll be given some 𝑦-value (called 𝑁), and you’ll be able to find some 𝑥-value
(called 𝑐). Note, there could be more than one possible value for the 𝑐.
This statement easiest understood through pictures:
𝑓(𝑏)
𝑁
𝑓(𝑎)
𝑎
𝑐
Given an 𝑁, we find a 𝑐.
𝑏
Or we can have:
𝑓(𝑎)
𝑁
𝑓(𝑏)
𝑎 𝑐1
Here, given an 𝑁, we can find 3 𝑐’s.
𝑐2
𝑐3
𝑏
The Intermediate Value Theorem
Example: Show that there is a root of the equation 𝑥 4 + 𝑥 − 3 = 0 on the interval [1,3].
Step 1: find the value 𝑁
When finding a root, by definition you want to find the 𝑥 that makes 𝑓(𝑥) = 0. So,
automatically, you get that 𝑁 = 0.
Step 2: find the equation 𝑓(𝑥)
𝑓(𝑥) is whatever is on the other side of the zero (ie: the 𝑁) in the given equation. Thus,
𝑓(𝑥) = 𝑥 4 + 𝑥 − 3
Step 3: verify that 𝑓(𝑎) and 𝑓(𝑏) are on either side of 𝑁
𝑓(𝑎) = 𝑓(1) = 14 + 1 − 3 = −1
𝑓(𝑏) = 𝑓(3) = 34 + 3 − 3 = 81
So, we get 𝑓(1) ≤ 0 ≤ 𝑓(3)
Step 4: conclusion
Hence, there is a 𝑥 = 𝑐 value in the interval [1,3] such that 𝑓(𝑐) = 0. In other words,
yes, there is a root of 𝑥 4 + 𝑥 − 3 = 0 in the interval [1,3].
Notice, we’re not finding the value of 𝑐, just that it exists.
𝜋
Example: Show that there is a root of the equation cos 𝑥 = 𝑥 on the interval �0, �.
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Step 1: find the value 𝑁
Finding a root means that 𝑁 = 0.
Step 2: find the function
cos 𝑥 = 𝑥 ⟹ cos 𝑥 − 𝑥 = 0
𝑓(𝑥) = cos 𝑥 − 𝑥
Step 3: verify that 𝑓(𝑎) and 𝑓(𝑏) are on either side of 𝑁
𝑓(𝑎) = 𝑓(0) = cos 0 − 0 = 1 − 0 = 1
𝜋
𝜋
𝜋
𝜋
𝜋
𝑓(𝑏) = 𝑓 � � = cos � � − = 0 − = −
2
2
2
2
2
𝜋
2
So, we get 𝑓(0) ≥ 0 ≥ 𝑓 � �
Step 4: conclusion
𝜋
Yes, there is a root of cos 𝑥 = 𝑥 on the interval �0, �.
2