If $f\left(x\right)\=2x\+1$, and $g\left(x\right)\=x^2\+x-1$, with all the reals as the common domain, then the rule for $h\left(x\right)\=f\left(x\right)\+g\left(x\right)$ is given by

$h\left(x\right)\=x^2\+3x$ 

and its graph is given in Figure 5.1.1.

The domain for $h\left(x\right)$ is again the set of all real numbers, while the range is the set of real numbers $y$ for which $y\>-\frac{9}{4}$.  One way to determine the minimum value of $y$ in the range of $h\left(x\right)$ is to find the axis of symmetry for the parabola that is the graph of $h\left(x\right)$.   

This axis of symmetry is midway between the $x$-intercepts, namely, $x\=0$ and $x\=-3$.  Hence, the axis of symmetry is $y\=-\frac{3}{2}$, and the vertex of the parabola is the point $\left(-\frac{3}{2}\,h\left(-\frac{3}{2}\right)\right)\=\left(-\frac{3}{2}\,-\frac{9}{4}\right)$.

Since $f\left(3\right)\=7$ and $g\left(3\right)\=11$, we have $h\left(3\right)\=18$ = $7\+11$, so $h\left(3\right)\=f\left(3\right)\+g\left(3\right)$.

The arithmetic sum of the functions

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

and

$g\left(x\right)\=x^2\+x-1$

namely

$h\left(x\right)\=f\left(x\right)\+g\left(x\right)$  

as well as its graph, its domain and range, and the function values $h\left(3\right)\,f\left(3\right)$, and $g\left(3\right)$ are provided by the Arithmetic of Functions  tutor.

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 5.1.2.

Note: When entering an expression in a tutor window, use * for multiplication.  Thus, for this example f(x) is entered as 2*x+1.

The rule for the sum is

$h\left(x\right)\=x^2\+3x$ 

From the graph of $h\left(x\right)$, we deduce that the domain is the set of all real numbers, and the range is the set of real numbers $y$ greater than or equal to $-\frac{9}{4}$.  One way to determine the minimum value of $y$ in the range of $h\left(x\right)$ is to find the axis of symmetry for the parabola that is the graph of $h\left(x\right)$.  This axis of symmetry is midway between the $x$-intercepts, namely, $x\=0$ and $x\=-3$.  Hence, the axis of symmetry is $y\=-\frac{3}{2}$, and the vertex of the parabola is the point $\left(-\frac{3}{2}\,h\left(-\frac{3}{2}\right)\right)\=\left(-\frac{3}{2}\,-\frac{9}{4}\right)$.

We can also see that because $f\left(3\right)\=7$ and $g\left(3\right)\=11$,

$h\left(3\right)\=18$ = $7\+11$ 

To launch the Arithmetic of Functions Tutor, click the following link: Arithmetic of Functions Tutor  

Enter the given data

Graph $h\left(x\right)$ 

RANGE: Determine the minimum of $h\left(x\right)$ 

Show $h\left(3\right)\=f\left(3\right)\+g\left(3\right)$ 

Enter the functions $f\left(x\right)$ and $g\left(x\right)$.

Define the function $h\left(x\right)\=f\left(x\right)\+g\left(x\right)$.

Graph $h\left(x\right)$.

Obtain the $x$-intercepts of $h\left(x\right)$.

Obtain the vertex of the parabola represented by $h\left(x\right)$ by evaluating $h\left(x\right)$ at the $x$-coordinate midway between the two $x$-intercepts,

From the graph of $h\left(x\right)$, it should be clear that the domain is the set of all real numbers, and the range is the set of all reals from the vertex of the parabola up.  Thus, the range is the set of reals $y$ for which $y\ge -\frac{9}{4}$.

Obtain the value $h\left(3\right)$.

Compute both $f\left(3\right)$ and $g\left(3\right)$.

Show that the sum of $f\left(3\right)$ and $g\left(3\right)$ is $h\left(3\right)$.

If $f\left(x\right)\=2x\+1$, and $g\left(x\right)\=x^2\+x-1$, with all the reals as the common domain, then the rule for $h\left(x\right)\=f\left(x\right)-g\left(x\right)$ is given by

$h\left(x\right)\=2\+x-x^2$ 

and its graph is given in Figure 5.2.1.

The domain for $h\left(x\right)$ is again the set of all real numbers, while the range is the set of real numbers $y$ for which $y\le \frac{9}{4}$ .  One way to determine the maximum value of $y$ in the range of $h\left(x\right)$ is to find the axis of symmetry for the parabola that is the graph of $h\left(x\right)$.

This axis of symmetry is midway between the $x$-intercepts, namely, $x\=-1$ and $x\=2$.  Hence, the axis of symmetry is $y\=\frac{1}{2}$, and the vertex of the parabola is the point $\left(\frac{1}{2}\,h\left(\frac{1}{2}\right)\right)\=\left(\frac{1}{2}\,\frac{9}{4}\right)$.

Since $f\left(3\right)\=7$ and $g\left(3\right)\=11$, we have $h\left(3\right)\=-4$ = $7-11$, so $h\left(3\right)\=f\left(3\right)-g\left(3\right)$.

The arithmetic difference of the functions

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

and

$g\left(x\right)\=x^2\+x-1$

namely,

$h\left(x\right)\=f\left(x\right)-g\left(x\right)$

as well as its graph, its domain and range, and the function values $h\left(3\right)\,f\left(3\right)$, and $g\left(3\right)$ are provided by the Arithmetic of Functions  tutor.

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 5.2.2.

The rule for the difference is

$h\left(x\right)\=-x^2\+x\+2$ 

From the graph of $h\left(x\right)$, we deduce that the domain is the set of all real numbers, while the range is the set of real numbers $y$ for which $y\le \frac{9}{4}$.  One way to determine the maximum value of $y$ in the range of $h\left(x\right)$ is to find the axis of symmetry for the parabola that is the graph of $h\left(x\right)$.  This axis of symmetry is midway between the $x$-intercepts, namely, $x\=-1$ and $x\=2$.  Hence, the axis of symmetry is $y\=\frac{1}{2}$, and the vertex of the parabola is the point $\left(\frac{1}{2}\,h\left(\frac{1}{2}\right)\right)\=\left(\frac{1}{2}\,\frac{9}{4}\right)$.

We can also see that because $f\left(3\right)\=7$ and $g\left(3\right)\=11$,

$h\left(3\right)\=18$ = $7-11$

To launch the Arithmetic of Functions Tutor, click the following link: Arithmetic of Functions Tutor  

Enter the given data

Graph $h\left(x\right)$ 

RANGE: Determine the maximum of $h\left(x\right)$ 

Show $h\left(3\right)\=f\left(3\right)-g\left(3\right)$ 

Enter the functions $f\left(x\right)$ and $g\left(x\right)$.

Define the function $h\left(x\right)\=f\left(x\right)- g\left(x\right)$.

Graph $h\left(x\right)$.

Obtain the $x$-intercepts of $h\left(x\right)$.

Obtain the vertex of the parabola represented by $h\left(x\right)$ by evaluating $h\left(x\right)$ at the $x$-coordinate midway between the two $x$-intercepts,

From the graph of $h\left(x\right)$, it should be clear that the domain is the set of all real numbers, and the range is the set of all reals from the vertex of the parabola down.  Thus, the range is the set of reals $y$ for which $y\le \frac{9}{4}$.

Obtain the value $h\left(3\right)$.

Compute both $f\left(3\right)$ and $g\left(3\right)$.

Show that the difference of $f\left(3\right)$ and $g\left(3\right)$ is $h\left(3\right)$.

If $f\left(x\right)\=2x\+1$, and $g\left(x\right)\=x^2\+x-1$, with all the reals as the common domain, then the rule for $h\left(x\right)\=f\left(x\right)g\left(x\right)$ is given by

$h\left(x\right)\=\left(2x\+1\right)\left(x^2\+x-1\right)$ = $2x^3\+3x^2-x-1$

and its graph is given in Figure 5.3.1.

The domain for $h\left(x\right)$ is again the set of all real numbers, as is the range.  

Since $f\left(3\right)\=7$ and $g\left(3\right)\=11$, we have $h\left(3\right)\=77$ = $\left(7\right)\left(11\right)$, so $h\left(3\right)\=f\left(3\right)g\left(3\right)$.

The arithmetic product of the functions

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

and

$g\left(x\right)\=x^2\+x-1$

namely,

$h\left(x\right)\=f\left(x\right)g\left(x\right)$ 

as well as its graph, its domain and range, and the function values $h\left(3\right)\,f\left(3\right)$, and $g\left(3\right)$ are provided by the Arithmetic of Functions  tutor.

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 5.3.2.

The rule for the product is

$h\left(x\right)\=\left(2x\+1\right)\left(x^2\+x-1\right)$

From the graph of $h\left(x\right)$, we deduce that the domain and range are both the set of all real numbers,

We can also see that because $f\left(3\right)\=7$ and $g\left(3\right)\=11$,

$h\left(3\right)\=77$ = $\left(7\right)\left(11\right)$  

To launch the Arithmetic of Functions Tutor, click the following link: Arithmetic of Functions Tutor  

Enter the given data

Graph $h\left(x\right)$ 

Domain and range of $h\left(x\right)$

Show $h\left(3\right)\=f\left(3\right) g\left(3\right)$ 

Enter the functions $f\left(x\right)$ and $g\left(x\right)$.

Define the function $h\left(x\right)\=f\left(x\right) g\left(x\right)$.

Graph $h\left(x\right)$.

From the graph of $h\left(x\right)$, it should be clear that both the domain and the range will be the set of all real numbers.

Obtain the value $h\left(3\right)$.

Compute both $f\left(3\right)$ and $g\left(3\right)$.

Show that the product of $f\left(3\right)$ and $g\left(3\right)$ is $h\left(3\right)$.

If $f\left(x\right)\=2x\+1$, and $g\left(x\right)\=x^2\+x-1$, with all the reals as the common domain, then the rule for $h\left(x\right)\=\frac{f\left(x\right)}{g\left(x\right)}$ is given by

$h\left(x\right)\=\frac{2x\+1}{x^2\+x-1}$

and its graph is given in Figure 5.4.1.

The domain for $h\left(x\right)$ is the set of all real numbers, less the two $x$-coordinates at which $h\left(x\right)$ has vertical asymptotes.  These two values, the zeros of $g\left(x\right)$, the denominator of $h\left(x\right)$, are $x\= -\frac{1}{2}\pm \frac{\sqrt{5}}{2}$, obtained with the quadratic formula.

The range is the set of all real numbers.

Since $f\left(3\right)\=7$ and $g\left(3\right)\=11$, we have $h\left(3\right)\=\frac{7}{11}$  so $h\left(3\right)\=\frac{f\left(3\right)}{g\left(3\right)}$.

The arithmetic quotient of the functions

$f\left(x\right)\=2x\+1$ 

and

$g\left(x\right)\=x^2\+x-1$ 

namely,

$h\left(x\right)\=\frac{f\left(x\right)}{g\left(x\right)}$ 

s well as its graph, its domain and range, and the function values $h\left(3\right)\,f\left(3\right)$, and $g\left(3\right)$ are provided by the Arithmetic of Functions  tutor.

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 5.4.2.

The rule for the quotient is

$h\left(x\right)\=\frac{2x\+1}{x^2\+x-1}$

From the graph of $h\left(x\right)$, we deduce that its domain is the set of all real numbers, less the two $x$-coordinates at which it has vertical asymptotes.  These two values, the zeros of $g\left(x\right)$, the denominator of $h\left(x\right)$, are $x\=-\frac{1}{2}\pm \frac{\sqrt{5}}{2}$, obtained with the quadratic formula.

We can also see that because $f\left(3\right)\=7$ and $g\left(3\right)\=11$,

$h\left(3\right)\=\frac{7}{11}$

To launch the Arithmetic of Functions Tutor, click the following link: Arithmetic of Functions Tutor  

Enter the given data

Graph $h\left(x\right)$ 

Domain of $h\left(x\right)$ 

Show $h\left(3\right)\=\frac{f\left(3\right)}{g\left(3\right)}$ 

Enter the functions $f\left(x\right)$ and $g\left(x\right)$.

Define the function $h\left(x\right)\=\frac{f\left(x\right)}{g\left(x\right)}$.

Graph $h\left(x\right)$.

Vertical asymptotes occur where $g\left(x\right)\=0$.

From the graph of $h\left(x\right)$, it should be clear that the domain is the set of all real numbers except for the two numbers at which $g\left(x\right)\=0$, namely, except for $x\=-\frac{1}{2}\pm \frac{\sqrt{5}}{2}$.  The range consists of all reals.

Obtain the value $h\left(3\right)$.

Compute both $f\left(3\right)$ and $g\left(3\right)$.