- Topic
- Continuity
- Difficulty
- Challenging
- Suggested time
- 5 min
- Marks
- 3
Continuity practice question 24
The join of the function below sits at , and the constant also appears in the first piece. Find every value of that makes continuous everywhere.
Both pieces are polynomials, so continuity can fail only at the join . Find the value of each piece there, in terms of .
Continuity needs , that is . Factor it rather than dividing by .
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
The join moves with c, but the test is the same wherever it is: the two pieces must agree at x = c. That gives a cubic in c. Factor it, keep every root, and check each one.
Working
The first piece gives the value at the join, and the second gives the limit from the right.
Bring everything to one side and factor. Dividing by c would assume c ≠ 0.
Each factor gives a root.
Check: at each join the two pieces meet. For c = 0, f is 0 for x ≤ 0 and 9x for x > 0.
Answer
The answer is c is −3, 0 or 3.
Check
At each of c = −3, 0 and 3 the two pieces agree at the join, at heights −27, 0 and 27.
Where this goes wrong
Dividing c³ = 9c by c, which loses c = 0, or taking only the positive square root of 9, which loses c = −3. Factor instead: c(c − 3)(c + 3) = 0 has three roots, and each one makes f continuous.
Leans on
Factoring a polynomial, from Functions, graphs and inverses
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