Chapter 1, Problem 13RE

Solution

To find: The domain and range of a function.

The domain and rage of the function $g\left(x\right)=x^2+2x+1$ are $\left(-\infty ,\infty \right)$ and $\left(0,\infty \right)$ respectively.

Given information: The function is $g\left(x\right)=x^2+2x+1$ .

Method/Formula used:

Suppose $y=f\left(x\right)$ be a function $f:A\to B$ , then the set A is called the domain of function f and set B the co-domain. Thus, the domain of a function $f\left(x\right)$ are the values of x for which the function $f\left(x\right)$ is defined. The range of the function is the set of points which are images of points of domain of function f . Thus, the range is a subset of the co-domain of the function f . The figure below explains domain, co-domain and the range of f .

  

Calculations:

The given function is $g\left(x\right)=x^2+2x+1$ . Let $g:A\to B$

(a) Domain:

Clearly, the function $g\left(x\right)=x^2+2x+1$ is defined for all real numbers, therefore, the set of all real numbers $ℝ\text{or}\left(-\infty ,\infty \right)$ is the domain of function $g\left(x\right)=x^2+2x+1$ .

(b) Range:

Suppose $y=g\left(x\right)$ , then the set of all y for which there exists an ‘ x’ in the domain of the function of $g\left(x\right)$ or set of points y for which the inverse function $g^{-1}\left(y\right)=x$ is defined, is the range of $g\left(x\right)$ .

Now let,

  $\left.\begin{array}{c}y=x^2+2x+1\\ ={\left(x+1\right)}^2\\ \Rightarrow x+1=\sqrt{y}\\ \Rightarrow x=\sqrt{y}-1\end{array}\right\}$ … (1)

From result (1), it is clear that for a negative $y\in B$ , there is no x in the set A . Therefore, corresponding to only positive numbers (y) of the set B , there exist numbers ( x ) in the set A . Therefore, the range of the function $g\left(x\right)$ is the set.

Hence, the range of the function $g\left(x\right)=x^2+2x+1$ is the interval $\left(0,\infty \right)$ .

Thus, the domain and rage of the function $g\left(x\right)=x^2+2x+1$ are $\left(-\infty ,\infty \right)$ and $\left(0,\infty \right)$ respectively.