Chapter 1.3, Problem 34E

To determine

To Show: The given function is continuous by using theorem $7$ .

Answer to Problem 34E

The domain of both the expression $f\left(x\right)=x^2+1$ and $g\left(x\right)=\sin x$ contains all real numbers, so, it is a continuous function.

Explanation of Solution

Given:

  $f\left(x\right)=\sin \left(x^2+1\right)$

Calculation:

Given that, $f\left(x\right)=\sin \left(x^2+1\right)$

Theorem 7:

The composite function $g°f\left(x\right)$ is continuous if $f\left(x\right)$ is continuous at $c$ and $g\left(x\right)$ is continuous at $f\left(c\right)$ .

The function $f\left(x\right)=\sin \left(x^2+1\right)$ is a composite function where $f\left(x\right)=x^2+1$ and $g\left(x\right)=\sin x$ .

Show$f\left(x\right)=x^2+1$is continuous:

Here $f\left(x\right)=x^2+1$ is a polynomial function. Generally, polynomial function is continuous everywhere and its domain is $R.$

Since the domain is all real numbers, $f\left(x\right)=x^2+1$ is continuous over all real numbers.

Show$g\left(x\right)=\sin x$

is continuous:

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation: $\left(-\infty ,\infty \right)$

Set-Builder Notation: $\left\{x|x\in ℝ\right\}$

Since the domain is all real numbers, $g\left(x\right)=\sin \left(x\right)$ is continuous over all real numbers $R$ .