Q. 92 - Calculus

Question

Consider the piecewise-defined function $f\left(x\right)=\left\{\begin{array}{ll}g\left(x\right),& if x\le c\\ h\left(x\right),& if x>c\end{array}\right.$.

Prove that if $g\left(x\right),h\left(x\right)$ are continuous and differentiable at $x=c$, and if $$\begin{gathered} g\left(c\right)=h\left(c\right), \\ g\text{'}\left(c\right)=h\text{'}\left(c\right) \end{gathered}$$, then f is differentiable at $x=c$.

Step-by-Step Solution

Answer

To prove that $g\left(x\right),h\left(x\right)$are continuous and differentiable at  $x=c$, first take the left and right hand limit of the function and simplify.

Then take the left and right hand derivative of the function and simplify.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}g\left(x\right),& if x\le c\\ h\left(x\right),& if x>c\end{array}\right. \\ g\left(c\right)=h\left(c\right), \\ g\text{'}\left(c\right)=h\text{'}\left(c\right) \end{gathered}$$

Step 2. Take the left and right hand limit of the function, followed by simplification.

The function is said to be continuous if the left hand limit, right hand limit and value of the function is same at that point.

The left hand limit of the function,

$$\begin{gathered} \underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{-}}{\lim }h\left(x\right) \\ =h\left(c\right) \end{gathered}$$

The right hand limit of the function,

$$\begin{gathered} \underset{x\to c^{+}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }h\left(x\right) \\ =g\left(c\right) \end{gathered}$$

Again at point $x=c$, the value of the function,

$f\left(c\right)=g\left(c\right)$

As $g\left(c\right)=h\left(c\right)$, therefore, the function is continuous at $x=c$.

Step 3. Take the left and right hand derivative of the function, followed by simplification.

As the function is continuous at $x=c$, therefore, the derivative test can be done at this point.

The left hand derivative of the function,

$$\begin{gathered} \underset{t\to 0^{-}}{\lim }\frac{f\left(c+h\right)-f\left(c\right)}{t}=\underset{t\to 0^{-}}{\lim }\frac{h\left(c+h\right)-h\left(c\right)}{t} \\ =h\text{'}\left(c\right) \end{gathered}$$

The right hand derivative of the function,

$$\begin{gathered} \underset{t\to 0^{+}}{\lim }\frac{f\left(c+h\right)-f\left(c\right)}{t}=\underset{t\to 0^{+}}{\lim }\frac{g\left(c+h\right)-g\left(c\right)}{t} \\ =g\text{'}\left(c\right) \end{gathered}$$

Since, $g\text{'}\left(c\right)=h\text{'}\left(c\right)$, then the function is differentiable at $x=c$.