Parameters for continuity and differentiability

Text

Find $a$ and $b$ so that $f(x) = \begin{cases}ax+1 & x\le 2 \\ x^2+b & x>2\end{cases}$ is continuous and differentiable at $x=2$.

Solution

Continuity at $x=2$. The value from the left is $f(2)=2a+1$; the limit from the right is $\lim_{x\to 2^+}(x^2+b)=4+b$. We need 2a+1 = 4+b. \tag{1}

Differentiability at $x=2$. The derivatives of the two pieces are $f'(x)=a$ for $x<2$ and $f'(x)=2x$ for $x>2$, whose right-hand limit is $2\cdot 2=4$. We need the slopes to coincide: a = 4. \tag{2}

Substituting $(2)$ into $(1)$: $\ 2(4)+1 = 4+b \Rightarrow 9 = 4+b \Rightarrow b=5$.

Therefore $\boxed{a=4,\ b=5}$. Check: from the left $f(2)=9$, from the right $4+5=9$ (continuous); slopes $4$ and $4$ (differentiable).

Links

Topics: Continuity
Concepts: Continuity · Differentiability · Piecewise-defined function
Skills: Reasoning by cases · Solving systems
Exercise types: Solving a system