- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
The definition of the derivative practice question 21
Find the values of the constants and for which the function below is differentiable everywhere, and enter .
Differentiable at 1 requires two things: continuous at 1, and matching one-sided derivatives at 1. That is two equations for the two unknowns.
The left piece has value 3 and derivative 4 at . The right piece has value and derivative .
An integer.
Theory UBC Math 100 §3.3CLP-1 §2.2Key results
Worked solution
Strategy
Away from x = 1 both pieces are polynomials, so only the join matters. At the join, continuity gives one equation and equal one-sided slopes give the other.
Working
The derivative of 2x² + 1 at 1, from the definition.
The limit exists only if the right piece also equals 3 at x = 1: that is continuity.
Matching slopes gives a; continuity gives b.
Answer
The answer is a is 4 and b is −1.
Check
With a = 4 and b = −1, 2x² + 1 − (4x − 1) = 2(x − 1)², so the line touches the parabola at x = 1 without crossing it.
Where this goes wrong
Matching the slopes but not the values, which gives a = 4 with b unconstrained. A function has to be continuous at a point before it can be differentiable there.
Leans on
Continuity at a join, from Continuity
Saved in this browser only.