- Topic
- Continuity
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Continuity practice question 15
The join of the function below sits at , where is a constant. Find every value of that makes continuous everywhere, and enter the sum of those values.
The join moves with . What must be true at for the two pieces to meet, and could more than one do it?
Both pieces are polynomials, so only the join at can break continuity. There the first piece gives , and the second gives the limit from the right, .
Set them equal and solve . How many roots does it have?
A fraction or a decimal.
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
Wherever the join is, both pieces are polynomials, so the only question is whether they meet at x = c. That is one equation in c, a quadratic, and every root gives a valid join.
Working
At the join the first piece gives the value, and the second gives the limit from the right.
Continuity at x = c: the two must agree.
Factor. Nothing restricts c, so both roots count.
Check each one: the pieces meet at the join.
Add the values.
Answer
The answer is c is 1/2 or 3, so the sum is 7/2.
Check
Without solving, the two roots of 2c² − 7c + 3 = 0 add to 7/2, minus the middle coefficient over the leading one.
Where this goes wrong
Stopping at the first root. The join moves with c, so the condition 2c² − 7c + 3 = 0 is a quadratic in c, and both of its roots make f continuous.
Leans on
Solving a quadratic by factoring, from Functions, graphs and inverses
Try a similar question
Find the value of the constant that makes continuous everywhere.Saved in this browser only.