Find a constant such that a given piecewise function is continuous everywhere

by RoRi

August 28, 2015

Given constants $a,b,c \in \mathbb{R}$ define:

$f(x) = \begin{cases} 2 \cos x & \text{if } x \leq c, \\ ax^2 + b & \text{if } x > c. \end{cases}$

If $b,c$ are fixed find all values of $a$ such that $f$ is continuous at $x = c$ .

By the definition of continuity, we know that $f$ continuous at $x = c$ means that $f(c)$ is defined and $\lim_{x \to c} f(x) = f(c)$ . Since $2 \cos x$ and $ax^2 + b$ are defined for all $x \in \mathbb{R}$ , we know that $f$ is defined for every $c \in \mathbb{R}$ . Then, to show that it is continuous we must show

$\lim_{x \to c} f(x) = f(c).$

From the definition of $f$ we know

$f(c) = 2 \cos c.$

Then, taking the limit as $x$ approaches $c$ from the right (since $\lim_{x \to c^-} f(x) = f(c)$ since $2 \cos x$ is continuous and $f(x) = 2 \cos x$ as $x$ approaches $c$ from the left),

$\lim_{x \to c^+} f(x) = \lim_{x \to c^+} ax^2 + b = ac^2 + b.$

So, we must have

$ac^2 + b = 2 \cos c \quad \implies \quad a = \frac{2 \cos c - b}{c^2} \quad \text{if } c \neq 0.$

If $c = 0$ then,

$ac^2 + b = 2 \cos c \quad \implies \quad b= 2 \quad \text{and } a \text{ is arbitrary}.$

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