Chapter 2.4, Problem 31E

To determine

To show : fis continuousfunctionon $\left(-\infty ,\infty \right)$ .

Answer to Problem 31E

The function is equal to 1 at $x=1$

  $f\left(1\right)=\sqrt{1}=1$ , it is continuous everywhere.

Explanation of Solution

Given:

  $f\left(x\right)=\{\begin{array}{l}{x}^{2}ifx<1\\ \sqrt{x}ifx\ge 1\end{array}$ .

Calculation:

The given function is $f\left(x\right)=\{\begin{array}{l}{x}^{2}ifx<1\\ \sqrt{x}ifx\ge 1\end{array}$ .

  $f\left(x\right)=x^2$ for $\left(-\infty ,1\right)$ and polynomials are continuous, so f is continuous for $\left(-\infty ,1\right)$ .

  $f\left(x\right)=\sqrt{x}$ for $\left(-\infty ,1\right)$ and root function are continuous on their domain, so f is continuous for $\left(1,\infty \right)$

From the left,

  $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }{x}^2=1$

From the right,

  $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\sqrt{x}=1$

From both sides, the limit is the same,

Therefore, $\underset{x\to 1}{\lim }f\left(x\right)=1$

The function itself equal to 1 at $x=1$

  $f\left(1\right)=\sqrt{1}=1$

Hence, the function is continuous everywhere.