Chapter 2.6, Problem 40PPSE

Solution

a.

To compare the domain and range of the transformed functions with that of the parent function.

A vertical translation of $f\left(x\right)$ does not change its domain but it changes its range. If the function $g\left(x\right)$ is obtained by vertically translating $f\left(x\right)$ to the upper side by $k$ units, it range will be $\left[-2+k,4+k\right]$ .Likewise, if the function $g\left(x\right)$ is transformed to the lower side by $k$ units, it range will be $\left[-2-k,4-k\right]$ .

During reflection about $x$ -axis, the domain of the function does not change and its range changes from $[-2,4]$ to $\left[-4,2\right]$ .

A vertical stretch by a factor of $a$ changes the range of the function from $[-2,4]$ to $\left[a\cdot \left(-2\right),a\cdot \left(4\right)\right]$ and it does not change its domain in any manner.

A horizontal translation of $f\left(x\right)$ does not change its range but it changes its domain. If the function $m\left(x\right)$ is obtained by horizontally translating $f\left(x\right)$ to the left side by $k$ units, its domain will be $\left[1-k,6-k\right]$. Likewise, if the function $m\left(x\right)$ is obtained by transforming $f\left(x\right)$ to the lower side by $k$ units, it domain will be $\left[1+k,6+k\right]$ .

Given the graph of a function $f\left(x\right)$:

  

Concept Used:

Given the graph of a function $y=f\left(x\right)$ .

Horizontal Translation:

The transformation $y=f\left(x-k\right)$ translates $y=f\left(x\right)$ horizontally right by $k$ units if $k>0$ .

The transformation $y=f\left(x-k\right)$ translates $y=f\left(x\right)$ horizontally left by $k$ units if $k<0$ .

Vertical Stretch:

The transformation $y=af\left(x\right)$ stretches the graph of $y=f\left(x\right)$ vertically by a factor of $a$ when $a>1$ .

Vertical Translation:

The transformation $y=f\left(x\right)+k$ translates $y=f\left(x\right)$ vertically up by $k$ units if $k>0$ .

The transformation $y=f\left(x\right)+k$ translates $y=f\left(x\right)$ vertically down by $k$ units if $k<0$ .

Reflection about the $x$-axis:

The transformation $y=-f\left(x\right)$ reflects $y=f\left(x\right)$ about the $x$ -axis.

Calculation:

Observe that the function $f\left(x\right)$ is defined only for those $x$ so that $1\le x\le 6$ .

That is, the domain of the function $f\left(x\right)$ is $[1,6]$ .

Now, the function $f\left(x\right)$ seems to be taking all the values from $-2\text{to }4$ .

That is, the range of the function is $[-2,4]$ .

Translate the graph of $f\left(x\right)$vertically to obtain $g\left(x\right)$:

Translate the graph of $f\left(x\right)$ vertically by some units say $2$ .

  

Observe that the function $g\left(x\right)$ obtained by translating $f\left(x\right)$ to the upper side has the range $[0,6]=\left[-2+2,4+2\right]$. Also, the function $g\left(x\right)$ transformed to the lower side has the range $\left[-4,2\right]=\left[-2-2,4-2\right]$. Please be noted that all the functions has the same domain.

Thus, if the function $g\left(x\right)$ is obtained by translating $f\left(x\right)$ to the upper side by $k$ units, its range will be $\left[-2+k,4+k\right]$. Also, if the function $g\left(x\right)$ is obtained by translating $f\left(x\right)$ to the lower side by $k$ units, its range will be $\left[-2-k,4-k\right]$. Moreover, a vertical translation does not change the domain of the function.

Reflect the graph of $f\left(x\right)$about the $x$-axis to obtain $h\left(x\right)$:

  

Observe that the domain of the function does not change and that the range of the function changes from $[-2,4]$ to $\left[-4,2\right]$ .

Stretch the graph of $f\left(x\right)$vertically and obtain $k\left(x\right)$:

Choose the factor of stretch to be $2$ .

Stretch the graph of $f\left(x\right)$ vertically by the factor of $2$ .

  

Observe that the range of the transformed function is $\left[-4,8\right]=\left[2\cdot \left(-2\right),2\cdot \left(4\right)\right]$ where as the domain remains unchanged.

Thus, a vertical stretch by a factor of $a$ changes the range of the function from $[-2,4]$ to $\left[a\cdot \left(-2\right),a\cdot \left(4\right)\right]$ and it does not change the domain in any manner.

Horizontally translate the graph of $f\left(x\right)$to obtain $m\left(x\right)$:

Translate the graph of $f\left(x\right)$ horizontally by some units, say $2$ .

  

Observe that the function $m\left(x\right)$ transformed to the left side has the domain $[-1,4]=\left[1-2,6-2\right]$. Also, the function $m\left(x\right)$ transformed to the right side has the domain $[3,8]=\left[1+2,6+2\right]$. Please note that the ranges of the functions are all the same.

Thus, if the function $m\left(x\right)$ is obtained by horizontally translating $f\left(x\right)$ to the left side by $k$ units, its domain will be $\left[1-k,6-k\right]$. Also, if the function $m\left(x\right)$ is obtained by horizontally translating $f\left(x\right)$ to the right side by $k$ units, its domain will be $\left[1+k,6+k\right]$. Moreover, a horizontal translation does not change its range.

Conclusion:

A vertical translation of $f\left(x\right)$ does not change its domain but it changes its range. If the function $g\left(x\right)$ is obtained by vertically translating $f\left(x\right)$ to the upper side by $k$ units, it range will be $\left[-2+k,4+k\right]$ .Likewise, if the function $g\left(x\right)$ is transformed to the lower side by $k$ units, it range will be $\left[-2-k,4-k\right]$ .

During reflection about $x$ -axis, the domain of the function does not change and its range changes from $[-2,4]$ to $\left[-4,2\right]$ .

A vertical stretch by a factor of $a$ changes the range of the function from $[-2,4]$ to $\left[a\cdot \left(-2\right),a\cdot \left(4\right)\right]$ and it does not change its domain in any manner.

A horizontal translation of $f\left(x\right)$ does not change its range but it changes its domain. If the function $m\left(x\right)$ is obtained by horizontally translating $f\left(x\right)$ to the left side by $k$ units, its domain will be $\left[1-k,6-k\right]$. Likewise, if the function $m\left(x\right)$ is obtained by transforming $f\left(x\right)$ to the lower side by $k$ units, it domain will be $\left[1+k,6+k\right]$ .

b.

To determine whether the observations of the previous part are generally happening.

The observations seem in part a are globally visible.

Explanation:

The observations seen in part a will be visible of any function under such transformations. It will definitely be visible over functions having bounded domain and bounded range as seen earlier.

If a function has unbounded domain, reflection about $x$ -axis, vertical translation and vertical stretch will not change its domain. Instead, it changes its range just like what observed in part a. In the case of horizontal translation, it would have changed the domain as per the observation. But, it will not happen since unbounded domain is not free to manipulate.

In the case of functions having unbounded range and bounded domain, horizontal shift will change the domain as discussed in part a. But, reflection about $x$ -axis, vertical translation and vertical stretches won’t make any change only since unbounded range is not that much flexible to manipulate.

Thus, the observations seem in part a are globally visible.

Conclusion:

The observations seem in part a are globally visible.