Rule

Domain

Range

$f\left(x\right)\=\sqrt{x\+1}$  

 { $x\in \mathrm{\Re }$, $x\ge -1$ }

 { $y\in \mathrm{\Re }$, $y\ge 0$ }

$g\left(x\right)\=\frac{x}{x^2\+1}$  

 { $x\in \mathrm{\Re }$ }

 { $y\in \mathrm{\Re }$, $-\frac{1}{2}\le y$$\le \frac{1}{2}$ }

Table 6.1.1   Rules $f\left(x\right)$ and $g\left(x\right)$, along with domains and ranges

Figure 6.1.1(a)   Graph of $f\left(x\right)\=\sqrt{x\+1}$

Figure 6.1.1(b)   Graph of $g\left(x\right)\=\frac{x}{x^2\+1}$

$x$

$0.8$

$0.9$

$1.0$

$1.1$

$1.2$

$g\left(x\right)$

$0.488$

$0.497$

$0.500$

$0.498$

$0.492$

Table 6.1.2   Values of $g\left(x\right)\=\frac{x}{x^2\+1}$

The rule for the composition $f\circ g$ is

$f\left(g\left(x\right)\right)\=\sqrt{\frac{x}{x^2\+1}\+1}$ 

At $x\=1$ we have

 $f\left(g\left(1\right)\right)\=\sqrt{\frac{1}{2}\+1}$ = $\sqrt{\frac{3}{2}}\=\frac{\sqrt{3}}{\sqrt{2}}$ = $\frac{\sqrt{6}}{2}$

A graph of this composition appears in Figure 6.1.2.

The minimum and maximum values of the fraction $g\left(x\right)\=\frac{x}{x^2\+1}$ are $-\frac{1}{2}$ and $\frac{1}{2}$, respectively.

That implies the minimum and maximum values of $f\left(g\left(x\right)\right)$ will be $\sqrt{1\/2}$ and $\sqrt{3\/2}$, respectively.   

Therefore, $g\left(x\right)$ provides to $f\left(x\right)$ only values in the domain of $f\left(x\right)$, so there are no values in the domain of $g\left(x\right)$ that cannot be passed along to $f\left(x\right)$.  Consequently, Figure 6.1.2 correctly suggests the domain and range of the composition $f\left(g\left(x\right)\right)$ to be respectively, the reals, and the interval $\left[\sqrt{1\/2}\,\sqrt{3\/2}\right]$.

The rule for the composition $g\circ f$ is

 $g\left(f\left(x\right)\right)\=\frac{\sqrt{x\+1}}{x\+2}$ 

At $x\=1$ we have $g\left(f\left(1\right)\right)\=\frac{\sqrt{2}}{3}$.

A graph of this composition appears in Figure 6.1.3.

This figure suggests the  domain for the composition $g\left(f\left(x\right)\right)$ is the set of real numbers $x$, where $x\>-1$, whereas the range is the interval $\left[0\,1\/2\right]$.

The domain of $f\left(x\right)$ is the set of real numbers $x$ for which $x\>-1$, while the range of $f\left(x\right)$ is the set of nonnegative reals.

If $g\left(x\right)$ is given such a nonnegative real number, it will produce a nonnegative real number no greater than $\frac{1}{2}$.  

Consequently, Figure 6.1.3 suggests the correct domain and range for the composition $g\left(f\left(x\right)\right)$.

Rule

Domain

Range

$f\left(x\right)\=\sqrt{x\+1}$  

 { $x\in \mathrm{\Re }$, $x\ge -1$ }

 { $y\in \mathrm{\Re }$, $y\ge 0$ }

$g\left(x\right)\=\frac{x}{x^2\+1}$  

 { $x\in \mathrm{\Re }$ }

 { $y\in \mathrm{\Re }$, $-\frac{1}{2}\le y$$\le \frac{1}{2}$ }

Table 6.1.1   Rules $f\left(x\right)$ and $g\left(x\right)$, along with domains and ranges

The domain and range for $f\left(x\right)$ can de deduced from the properties of the square-root function, or from a graph.  The domain for $g\left(x\right)$ can be deduced from its rule, but the range can only be approximated from its graph.  Graphs of $f\left(x\right)$ and $g\left(x\right)$ can be obtained with the Composition Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 6.1.4.

This tutor provides a graph of $f\left(x\right)\,g\left(x\right)$, and a choice of the compositions $f\left(g\left(x\right)\right)$ or $g\left(f\left(x\right)\right)$.

The determination of the exact height to which a function such as $g\left(x\right)$ rises is difficult with just the tools of algebra.

To launch the Composition Tutor, click the following link: Composition Tutor  

The Composition Tutor  will provide $f\left(g\left(x\right)\right)$, the rule for the composition $f\circ g$.

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 6.1.5.

In addition to graphs of $f\left(x\right)\,g\left(x\right)$, and $f\left(g\left(x\right)\right)$, the values of these three functions at $x\=1$ are provided.

This tutor will show that the rule for the composition $f\circ g$ is

$f\left(g\left(x\right)\right)\=\sqrt{\frac{x}{x^2\+1}\+1}$  =  $\sqrt{\frac{x^2\+x\+1}{x^2\+1}}$

It will also show that at $x\=1$ we have

$f\left(g\left(1\right)\right)\=\sqrt{\frac{1}{2}\+1}$ = $\sqrt{\frac{3}{2}}\=\frac{\sqrt{3}}{\sqrt{2}}$ = $\frac{\sqrt{6}}{2}$ = $\frac{\sqrt{2}}{2}$ $\sqrt{3}$ 

For this composition, the Composition Tutor also provides a graph similar to the one in Figure 6.1.6, a reproduction of Figure 6.1.2.

The minimum and maximum values of the fraction $g\left(x\right)\=\frac{x}{x^2\+1}$ are $-\frac{1}{2}$ and $\frac{1}{2}$, respectively.  That implies the minimum and maximum values of $f\left(g\left(x\right)\right)$ will be $\sqrt{1\/2}$ and $\sqrt{3\/2}$, respectively. Therefore, $g\left(x\right)$ provides to $f\left(x\right)$ only values in the domain of $f\left(x\right)$, so there are no values in the domain of $g\left(x\right)$ that cannot be passed along to $f\left(x\right)$.  

Consequently, Figure 6.1.2 correctly suggests the domain and range of the composition $f\left(g\left(x\right)\right)$ to be respectively, the reals, and the interval $\left[\sqrt{1\/2}\,\sqrt{3\/2}\right]$.

The Composition Tutor  will provide $g\left(f\left(x\right)\right)$, the rule for the composition $g\circ f$. (Clicking this link will launch the tutor with the solution embedded as shown in the  thumbnail image at the right.)  In addition to graphs of $f\left(x\right)\,g\left(x\right)$, and $g\left(f\left(x\right)\right)$, the values of these three functions at $x\=1$ are provided.

This tutor will show that the rule for the composition $g\circ f$ is

$g\left(f\left(x\right)\right)\=\frac{\sqrt{x\+1}}{x\+2}$ 

It will also show that at $x\=1$ we have

$g\left(f\left(1\right)\right)\=\frac{\sqrt{2}}{3}$ 

For this composition, the Composition Tutor also provides a graph similar to the one in Figure 6.1.7, a reproduction of Figure 6.1.3.

This figure suggests the domain for the composition $g\left(f\left(x\right)\right)$ is the set of real numbers $x$, where $x\>-1$, whereas the range is the interval $\left[0\,1\/2\right]$.

The domain of $f\left(x\right)$ is the set of real numbers $x$ for which $x\>-1$, while the range of $f\left(x\right)$ is the set of nonnegative reals.  If $g\left(x\right)$ is given such a nonnegative real number, it will produce a nonnegative real number no greater than $\frac{1}{2}$.  Consequently, the figure above suggests the correct domain and range for the composition $g\left(f\left(x\right)\right)$.

To launch the Composition Tutor, click the following link: Composition Tutor  

The domain and range of $f\left(x\right)$ can be inferred from its graph.

The domain is the set of all real numbers $x$ satisfying $x\ge -1$, while the range is the set of all real numbers $y$ satisfying $y\ge 0$.

The domain and range of $g\left(x\right)$ can be inferred from its graph.

It appears that the domain of $g\left(x\right)$ is the set of all real numbers, but the range is the closed interval $\left[-1\/2\,1\/2\right]$.  The graph suggests the lowest and highest points on the graph of $g\left(x\right)$ are $\left(-1\,-\frac{1}{2}\right)$ and $\left(1\,\frac{1}{2}\right)$, respectively.  This hypothesis is confirmed by the following calculations.

Enter the given functions

Obtain and graph the composition $f\left(g\left(x\right)\right)$ 

Evaluations at $x\=a\=1$ 

Domain and Range of the composition $f\left(g\left(x\right)\right)$ 

Figure 6.1.2 suggests the domain for the composition $f\left(g\left(x\right)\right)$ is the set of real numbers, whereas the range is a bounded set localized about $y\=1$.  The best tools for determining the maximum and minimum values of a function are developed in the calculus.  Here, however, we can do the following.

The minimum and maximum values of the fraction $g\left(x\right)\=\frac{x}{x^2\+1}$ are $-1\/2$ and $1\/2$, respectively.  That implies the minimum and maximum values of $f\left(g\left(x\right)\right)$ will be $\sqrt{1\/2}$ and $\sqrt{3\/2}$, respectively.

Therefore, $g\left(x\right)$ provides to $f\left(x\right)$ only values in the domain of $f\left(x\right)$, so there are no values in the domain of $g\left(x\right)$ that cannot be passed along to $f\left(x\right)$.  Consequently, Figure 6.1.2 correctly suggests the domain and range of the composition $f\left(g\left(x\right)\right)$ to be respectively, the reals, and the interval $\left[\sqrt{1\/2}\,\sqrt{3\/2}\right]$.

The extreme values of $g\left(x\right)$ can be obtained numerically as follows.

Enter the given functions

Obtain and graph the composition $g\left(f\left(x\right)\right)$

Evaluations at $x\=a\=1$ 

Domain and Range of the composition $g\left(f\left(x\right)\right)$

Figure 6.1.3 suggests the domain for the composition $g\left(f\left(x\right)\right)$ is the set of real numbers $x$, where $x\ge -1$, whereas the range is the interval $\left[0\,1\/2\right]$.

The domain of $f\left(x\right)$ is the set of real numbers $x$ for which $x\ge -1$, while the range of $f\left(x\right)$ is the set of nonnegative reals.  

If $g\left(x\right)$ is given such a nonnegative real number, it will produce a nonnegative real number no greater than $1\/2$.  

Consequently, Figure 6.1.3 suggests the correct domain and range for the composition $g\left(f\left(x\right)\right)$.

The maximum value of the composition $g\left(f\left(x\right)\right)$ can be determined numerically via the Optimization Assistant.

Enter the functions $f$ and $g$.

Graph $f$ as per Figure 6.1.1(a).

From this graph, or from the rule for $f$, infer that the domain of $f$ is the interval $\left[-1\,\infty \right)$, and that the range is the interval $\left[0\,\infty \right)\.$

Graph $g$ as per Figure 6.1.1(b).

That the domain of $g$ is the set of all real numbers can be inferred from Figure 6.1.1(b).  It is more difficult to determine that the range is the interval $\left[-1\/2\,1\/2\right]$.

Obtain the minimum and maximum for $g$.  The minimize and maximize commands provide the extreme values as well as the $x$-coordinate where the extreme occurs.

Enter the functions $f$ and $g$.

Obtain the composition $f\left(g\left(x\right)\right)$.

Obtain $f\left(g\left(1\right)\right)$.

Obtain $g\left(1\right)$.

Obtain $f\left(\frac{1}{2}\right)\=f\left(g\left(1\right)\right)$.

Graph $f\left(g\left(x\right)\right)$, thereby obtaining Figure 6.1.2.

Figure 6.1.2 suggests the domain for the composition $f\left(g\left(x\right)\right)$ is the set of real numbers, whereas the range is a bounded set localized about $y\=1$.  The best tools for determining the maximum and minimum values of a function are developed in the calculus.  Here, however, we reason as follows.  

The minimum and maximum values of the fraction $g\left(x\right)\=\frac{x}{x^2\+1}$ are $-\frac{1}{2}$ and $\frac{1}{2}$, respectively.  That implies the minimum and maximum values of $f\left(g\left(x\right)\right)$ will be $\sqrt{1\/2}$ and $\sqrt{3\/2}$, respectively. 

Therefore, $g\left(x\right)$ provides to $f\left(x\right)$ only values in the domain of $f\left(x\right)$, so there are no values in the domain of $g\left(x\right)$ that cannot be passed along to $f\left(x\right)$.  Consequently, Figure 6.1.2 correctly suggests the domain and range of the composition $f\left(g\left(x\right)\right)$ to be respectively, the reals, and the interval $\left[\sqrt{1\/2}\,\sqrt{3\/2}\right]$.

Obtain the minimum and maximum for $f\left(g\left(x\right)\right)$.  The minimize and maximize commands provide the extreme values as well as the $x$-coordinate where the extreme occurs.

Enter the functions $f$ and $g$.

Obtain the composition $g\left(f\left(x\right)\right)$.

Compute $g\left(f\left(1\right)\right)$.

Obtain $f\left(1\right)$.

Obtain $g\left(\sqrt{2}\right)\=g\left(f\left(1\right)\right)$.

Graph $g\left(f\left(x\right)\right)$, thereby obtaining Figure 6.1.3.

Figure 6.1.3 suggests the domain for the composition $g\left(f\left(x\right)\right)$ is the set of real numbers $x$, where $x\ge -1$, whereas the range is the interval $\left[0\,1\/2\right]$.

The domain of $f\left(x\right)$ is the set of real numbers $x$ for which $x\ge -1$, while the range of $f\left(x\right)$ is the set of nonnegative reals.  

If $g\left(x\right)$ is given such a nonnegative real number, it will produce a nonnegative real number no greater than $1\/2$.  

Consequently, Figure 6.1.3 suggests the correct domain and range for the composition $g\left(f\left(x\right)\right)$.

Obtain the minimum and maximum for $f\left(g\left(x\right)\right)$.  The maximize command provides the extreme values as well as the $x$-coordinate where the extreme occurs.

Rule

Domain

Range

$f\left(x\right)\=\frac{x}{x+1}$  

 { $x\in \mathrm{\Re }$, $x\ne$$-1$ }

 { $y\in \mathrm{\Re }$, $y\ne 1$ }

$g\left(x\right)\=\frac{1}{x-1}$  

 { $x\in \mathrm{\ℜ}\,x\ne 1$ }

 { $y\in \mathrm{\Re }$, $y\ne 0$ }

Table 6.2.1   Rules $f\left(x\right)$ and $g\left(x\right)$, along with domains and ranges

The domain and range for $f\left(x\right)$ can be deduced from the properties of the fraction $\frac{x}{x\+1}$, or from the graph of $f\left(x\right)$ in Figure 6.2.1(a).  

The vertical asymptote whose equation is $x\=-1$, and the horizontal asymptote $y\=1$ determine the domain and range, respectively.

Consequently, the domain consists of all real numbers except the number $-1$, whereas the range consists of all real numbers except the number 1.

The domain and range for $g\left(x\right)$ can be deduced from the properties of the fraction $\frac{1}{x-1}$, or from the graph of $g\left(x\right)$ in Figure 6.2.1(b).  

The vertical asymptote whose equation is $x\=1$, and the horizontal asymptote $y\=0$ determine the domain and range, respectively.

Consequently, the domain consists of all real numbers except the number 1, whereas the range consists of all real numbers except the number 0.

The rule for the composition $f\circ g$ is

 $f\left(g\left(x\right)\right)\=\frac{\left(\frac{1}{x-1}\right)}{\frac{1}{x-1}\+1}$  =  $\frac{1}{1\+x-1}\=\frac{1}{x}$

At $x\=2$ we have $f\left(g\left(2\right)\right)\=\frac{1}{2}$.

A graph of this composition appears in Figure 6.2.2.

Figure 6.2.2 seems to suggest the domain for the composition $f\left(g\left(x\right)\right)$ is the set of real numbers $x$ satisfying $x\ne 0$, and the range is the set of real numbers $y$ satisfying $y\ne 0$.  However, this is naive.  The analysis must be more sophisticated.

The "inside" function $g\left(x\right)\=\frac{1}{x-1}$ cannot be given $x\=1$.  Hence, the domain of the composition $f\left(g\left(x\right)\right)$ cannot contain $x\=1$.  Moreover, the function $f\left(x\right)\=\frac{x}{x\+1}$ cannot be given $x\=-1$.  We must determine which value, if any, in the domain of $g\left(x\right)$, delivers to $f\left(x\right)$ the value $-1$.  Thus, we must solve the equation

$g\left(x\right)\=\frac{1}{x-1}$ = $-1$ 

obtaining

$1\=1-x$ 

from which it follows that $x\=0$.  Thus, the two real numbers $x\=0$ and $x\=1$, cannot be in the domain of the composition $f\left(g\left(x\right)\right)$.  Consequently, if $x\=1$ cannot be in the domain of the rule $f\left(g\left(x\right)\right)\=\frac{1}{x}$, then $y\=1$ will not be in the range of the composition.  Hence, the range of $f\left(g\left(x\right)\right)$ is the set of real numbers $y$, except for the two values $y\=0$, and $y\=1$.

The rule for the composition $g\circ f$ is

$g\left(f\left(x\right)\right)\=\frac{1}{\left(\frac{x}{x\+1}\right)-1}$  =  $\frac{x\+1}{x-x-1}\=-\left(x+1\right)$

At $x\=2$ we have $g\left(f\left(2\right)\right)\=-3$.

A graph of this composition appears in Figure 6.2.3.

It might appear from Figure 6.2.3 that both the domain and range for the composition $f\left(g\left(x\right)\right)\=-\left(x+1\right)$ would be the set of all real numbers.  However, that is naive.  A more sophisticated analysis is required.