Chapter 1, Problem 14RE

Solution

To find: The domain and range of a function.

The domain and rage of the function $h\left(x\right)={\left(x-2\right)}^2+5$ are $\left(-\infty ,\infty \right)$ and $\left[5,\infty \right)$ respectively.

Given information: The function is $h\left(x\right)={\left(x-2\right)}^2+5$ .

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. The range of the function is the set of points which are images of points of domain of function f . Thus, the range is the domain of the function $f^{-1}\left(x\right)$ . The figure below explains domain, co-domain and the range of f .

  

Calculations:

The given function is $h\left(x\right)={\left(x-2\right)}^2+5$ . Let $h:A\to B$

(a) Domain:

Clearly, the function $h\left(x\right)={\left(x-2\right)}^2+5$ 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 $h\left(x\right)={\left(x-2\right)}^2+5$ .

(b) Range:

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

Now let,

  $\left.\begin{array}{c}y={\left(x-2\right)}^2+5\\ \sqrt{y-5}=x-2\\ x\Rightarrow 2+\sqrt{y-5}\\ h^{-1}\left(y\right)\Rightarrow 2+\sqrt{y-5}\end{array}\right\}$ … (1)

From result (1), it is clear that $h^{-1}\left(y\right)$ is not defined when $y-5$ is negative. Thus, $h^{-1}\left(y\right)$ is defined only for $y\ge 5$ .

Hence, the range of the function $h\left(x\right)={\left(x-2\right)}^2+5$ is the set of real numbers greater than 5, i.e., the interval $\left[0,\infty \right)$ .

Thus, the domain and rage of the function $h\left(x\right)={\left(x-2\right)}^2+5$ are $\left(-\infty ,\infty \right)$ and $\left[5,\infty \right)$ respectively.