# Dilations Help (1 Viewer)

#### csi

##### Member
Hi guys,

I need some help on a couple of questions:

1. Find the equation if f(x)=1/3 x^3 is translated horizontally 1/2 units to the right.

Why is the answer 1/2(2x-1)^3 instead of 1/3(x-1/2)^3? What are the differences?

2. f(x)=ln x has a vertical translation of 2 upwards and vertical dilation with factor 1/2, a horizontally dilation with factor 3 and a horizontal translation of 1/5 to the right.

Up until the last transformation, I've got 1/2 ln 3x+2, how do I show the final horizontal translation of 1/5?

Thanks guys!

#### Qeru

##### Well-Known Member
Hi guys,

I need some help on a couple of questions:

1. Find the equation if f(x)=1/3 x^3 is translated horizontally 1/2 units to the right.

Why is the answer 1/2(2x-1)^3 instead of 1/3(x-1/2)^3? What are the differences?

2. f(x)=ln x has a vertical translation of 2 upwards and vertical dilation with factor 1/2, a horizontally dilation with factor 3 and a horizontal translation of 1/5 to the right.

Up until the last transformation, I've got 1/2 ln 3x+2, how do I show the final horizontal translation of 1/5?

Thanks guys!
What Textbook (let me guess MIF), your right textbook is wrong (unless you are missing some sort of dilation?) for 1 as demonstrated by this graph:

Red shows textbooks answers which is clearly dilated on top of being translated whereas blue shows your answer which is simply translated 1/2 to the right.

For b
To show the horizontal translation replace x with x-1/5 i.e.

#### csi

##### Member
What Textbook (let me guess MIF), your right textbook is wrong (unless you are missing some sort of dilation?) for 1 as demonstrated by this graph:

View attachment 30127
Red shows textbooks answers which is clearly dilated on top of being translated whereas blue shows your answer which is simply translated 1/2 to the right.

For b
To show the horizontal translation replace x with x-1/5 i.e. View attachment 30128
Thanks

#### CM_Tutor

##### Moderator
Moderator
Hi guys,

I need some help on a couple of questions:

1. Find the equation if f(x)=1/3 x^3 is translated horizontally 1/2 units to the right.

Why is the answer 1/2(2x-1)^3 instead of 1/3(x-1/2)^3? What are the differences?

2. f(x)=ln x has a vertical translation of 2 upwards and vertical dilation with factor 1/2, a horizontally dilation with factor 3 and a horizontal translation of 1/5 to the right.

Up until the last transformation, I've got 1/2 ln 3x+2, how do I show the final horizontal translation of 1/5?

Thanks guys!
@Qeru is right that the first one sounds like an error in the textbook, though I wonder if it is a typo as the translated form is

$\bg_white \frac{1}{3}\left(x-\frac{1}{2}\right)^3 = \frac{1}{3}\left(\frac{2x-1}{2}\right)^3 = \frac{1}{3}\times\frac{(2x-1)^3}{8} = \frac{1}{24}(2x-1)^3$

and so the "2" in the denominator of the textbook's answer could be a typo for "24"

On the second question, Qeru is right about the translation but I wonder if you have made a mistake with your "+2"?

\bg_white \begin{align*} \text{We begin with} \quad f(x) &= \ln{x} \\ \text{After } \color{blue} \text{vertical translation of 2 units upwards, } \color{black} \text{we get:} \quad f(x) &= \ln{x} \color{blue}+ 2\color{black} = 2 + \ln{x}\\ \text{After } \color{red} \text{vertical dilation by a factor of \frac{1}{2}, } \color{black} \text{we get:} \quad f(x) \color{red}\div \frac{1}{2}\color{black} &= 2 + \ln{x} \\ f(x) &= \color{red}\frac{1}{2}\color{black}\left(2 + \ln{x}\right) = \color{red}1\color{black} + \frac{1}{2}\ln{x} \\ \text{After } \color{blue} \text{horizontal dilation by a factor of 3, } \color{black} \text{we get:} \quad f(x) &= 1 + \frac{1}{2}\ln{\color{blue}3x\color{black}} \\ \text{After } \color{blue} \text{horizontal translation of \frac{1}{5} units to the right, } \color{black} \text{we get:} \quad f(x) &= 1 + \frac{1}{2}\ln{\left[3 \color{blue} \left(x-\frac{1}{5}\right) \color{black}\right]} \\ &= 1 + \frac{1}{2}\ln{\frac{3(5x - 1)}{5}} \end{align*}

Note that the log laws allow us to write this answer in a number of forms, including:

$\bg_white f(x) = 1 + \frac{1}{2}\ln{\frac{3}{5}} + \frac{1}{2}\ln{(5x - 1)} = \frac{1}{2}\ln{\frac{3e^2(5x - 1)}{5}}$

It is wise to consider the possibility that the textbook may be incorrect, especially with more recently published textbooks. I am keeping a list of errors in the Cambridge books, for example... but another possibility is that the textbook answer is written in a different form that is equivalent to your answer. In exams, unless the form in which the answer is required is specified, all reasonable equivalent answers should be accepted, so you don't need to stress if the textbook authors choose a different form for presenting the answer. Obviously, considering whether you have made a mistake is the first option, but if you can't see one, bear in mind the other possibilities... and good job on posting here to check.

PS: It is helpful to people like me to state where a question comes from. I am planning to send Cambridge a list of mistakes / typos / formatting issues etc that I find in their books, so seeing what are errors and where they are is helpful to me. Thanks.

#### csi

##### Member
@Qeru is right that the first one sounds like an error in the textbook, though I wonder if it is a typo as the translated form is

$\bg_white \frac{1}{3}\left(x-\frac{1}{2}\right)^3 = \frac{1}{3}\left(\frac{2x-1}{2}\right)^3 = \frac{1}{3}\times\frac{(2x-1)^3}{8} = \frac{1}{24}(2x-1)^3$

and so the "2" in the denominator of the textbook's answer could be a typo for "24"

On the second question, Qeru is right about the translation but I wonder if you have made a mistake with your "+2"?

\bg_white \begin{align*} \text{We begin with} \quad f(x) &= \ln{x} \\ \text{After } \color{blue} \text{vertical translation of 2 units upwards, } \color{black} \text{we get:} \quad f(x) &= \ln{x} \color{blue}+ 2\color{black} = 2 + \ln{x}\\ \text{After } \color{red} \text{vertical dilation by a factor of \frac{1}{2}, } \color{black} \text{we get:} \quad f(x) \color{red}\div \frac{1}{2}\color{black} &= 2 + \ln{x} \\ f(x) &= \color{red}\frac{1}{2}\color{black}\left(2 + \ln{x}\right) = \color{red}1\color{black} + \frac{1}{2}\ln{x} \\ \text{After } \color{blue} \text{horizontal dilation by a factor of 3, } \color{black} \text{we get:} \quad f(x) &= 1 + \frac{1}{2}\ln{\color{blue}3x\color{black}} \\ \text{After } \color{blue} \text{horizontal translation of \frac{1}{5} units to the right, } \color{black} \text{we get:} \quad f(x) &= 1 + \frac{1}{2}\ln{\left[3 \color{blue} \left(x-\frac{1}{5}\right) \color{black}\right]} \\ &= 1 + \frac{1}{2}\ln{\frac{3(5x - 1)}{5}} \end{align*}

Note that the log laws allow us to write this answer in a number of forms, including:

$\bg_white f(x) = 1 + \frac{1}{2}\ln{\frac{3}{5}} + \frac{1}{2}\ln{(5x - 1)} = \frac{1}{2}\ln{\frac{3e^2(5x - 1)}{5}}$

It is wise to consider the possibility that the textbook may be incorrect, especially with more recently published textbooks. I am keeping a list of errors in the Cambridge books, for example... but another possibility is that the textbook answer is written in a different form that is equivalent to your answer. In exams, unless the form in which the answer is required is specified, all reasonable equivalent answers should be accepted, so you don't need to stress if the textbook authors choose a different form for presenting the answer. Obviously, considering whether you have made a mistake is the first option, but if you can't see one, bear in mind the other possibilities... and good job on posting here to check.

PS: It is helpful to people like me to state where a question comes from. I am planning to send Cambridge a list of mistakes / typos / formatting issues etc that I find in their books, so seeing what are errors and where they are is helpful to me. Thanks.
Thanks for the explanation, but these questions aren't from textbooks