- Topic
- Continuity
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Continuity practice question 9
Let , so that and . Does the Intermediate Value Theorem guarantee a number in with ?
The theorem has a hypothesis as well as a conclusion. Before using the signs at the ends, check that is continuous at every point of .
Where in is undefined? And can a fraction whose numerator is ever equal ?
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
The Intermediate Value Theorem needs two things: f continuous on the closed interval, and a target value between the values at the ends. Check the first before using the second.
Working
The theorem, with its hypothesis.
So f is not continuous on [−1, 1], and the theorem says nothing here.
In fact there is no zero at all: the graph changes sign across its vertical asymptote at x = 0 without meeting the axis.
Answer
The answer is No, because f is not continuous on [−1, 1]; in fact 1/x is never 0.
Check
f(−0.01) = −100 and f(0.01) = 100: the values change sign across x = 0, where f is undefined, without ever being 0.
Where this goes wrong
Using the theorem on the signs at the ends alone. Without continuity on the whole interval a function can change sign without a zero, as 1/x does across its asymptote at 0.
Leans on
The domain of a rational function, from Functions, graphs and inverses
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