Determining the parameters for continuity

When a piecewise function contains parameters, imposing continuity (and possibly differentiability) at the junction point provides the equations that determine them.

Example — Determining the parameters for continuity

Let $f(x) = \begin{cases} x^2+bx & x<2 \\ \ln(2x+a) & x\ge 2\end{cases}$ Find $a,b$ so that $f$ is continuous and differentiable at $x=2$.

Continuity: $\lim_{x\to 2^-}(x^2+bx) = 4+2b$; $\ f(2) = \ln(4+a)$. We need: $4+2b = \ln(4+a).$

Differentiability: $f'(x) = \begin{cases}2x+b & x<2 \\ \dfrac{2}{2x+a} & x\ge 2\end{cases}$. The left limit of the derivative is $4+b$, the right one is $\dfrac{2}{4+a}$. We need: $4+b = \frac{2}{4+a}, \qquad\text{from which}\qquad 4+a = \frac{2}{4+b}.$

Substituting into the first condition: $4+2b = \ln\!\left(\frac{2}{4+b}\right).$ It is a transcendental equation: it is solved numerically (or by trial), not with elementary algebraic methods.

The example shows an important point: continuity and differentiability give two distinct conditions. Continuity links the values of the two formulae; differentiability links their slopes. Together they form a system that, in general, may even be transcendental.

Links

Topics: Continuita
Concepts: Continuita · Derivabilita · Funzione definita a tratti
Skills: Ragionare per casi · Risolvere equazioni