- Topic
- Continuity
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Continuity practice question 19
A function is continuous on , and the table gives some of its values. What is the least number of zeros must have in ?
By the Intermediate Value Theorem, a continuous has a zero between two neighbouring entries whose values have opposite signs. Where does the sign of change in the table?
One entry is already . Does the change in sign from to need a zero other than that one?
A whole number.
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
Each sign change between neighbouring entries forces a zero, by the Intermediate Value Theorem, and an entry of 0 is a zero itself. Count them without counting any zero twice; a function that is straight between the entries shows that no more are forced.
Working
f is continuous on [0, 1] and changes sign, so by the Intermediate Value Theorem it has a zero in (0, 1).
A zero in the table itself. The change from f(2) = −2 to f(4) = 2 is accounted for by this one.
Another sign change, so another zero, in (5, 6).
Joining the points of the table by straight lines gives a continuous f with exactly these three zeros, so no more are forced.
Answer
The answer is three zeros.
Check
The function that is straight between the entries crosses zero once in (0, 1), once at x = 3 and once in (5, 6), and nowhere else.
Where this goes wrong
Counting only the sign changes and missing the 0 in the table, or counting the change from x = 2 to x = 4 as another zero beside the one at x = 3.
Leans on
Reading a function from a table, from Functions, graphs and inverses
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