Chapter 3.5, Problem 52HP

To determine

To find: Whether $af\left(x\right)$ and $f\left(ax\right)$ are equivalent or not.

Answer to Problem 52HP

Yes, if $a=1$

No, if $a\ne 1$

Explanation of Solution

Given information:

  $af\left(x\right)$

  $f\left(ax\right)$

Concept:

A Translation is a movement of the graph either horizontally parallel to x-axis or vertically parallel to y-axis.

The graph of $y=f\left(x\right)$ where $f\left(x\right)=x$ is same as the graph of $y=x$ . Writing the equations as functions in the form of $f\left(x\right)$ is useful when applying translations and reflections to graphs.

If $g\left(x\right)=f\left(x\right)+k$ it is known as vertical shift.

If $k>0$ it moves upwards.

If $k<0$ it moves downwards.

If $g\left(x\right)=f\left(x+l\right)$ it is known as horizontal shift.

If $l<0$ moves it right.

If $l>0$ moves it left.

If $g\left(x\right)=-f\left(x\right)$ it is known as Vertical reflection it means reflects about X-axis

If $g\left(x\right)=f\left(-x\right)$ it is known as Horizontal reflection it means reflects about Y-axis

If $g\left(x\right)=af\left(x\right)$

If $a>1$ it is known as Vertical stretch.

If $0<a<1$ it is known as Vertical compression.

If $g\left(x\right)=f\left(bx\right)$

If $b>1$ it is known as Horizontal compression.

If $0<b<1$ it is known as Horizontal Stretch.

Calculation:

Consider $a=1$ ,

  $af\left(x\right)=1\times f\left(x\right)$

  $\Rightarrow af\left(x\right)=f\left(x\right)$

  $f\left(ax\right)=f\left(1\times x\right)$

  $\Rightarrow f\left(ax\right)=f\left(x\right)$

Therefore, Alex assumption is True, i.e. $af\left(x\right)$ and $f\left(ax\right)$ are equivalent.

Consider $a\ne 1$ ,

From the above concept of,

If $g\left(x\right)=af\left(x\right)$

If $a>1$ it is known as Vertical stretch.

If $0<a<1$ it is known as Vertical compression.

Therefore, the graph of the function $af\left(x\right)$ is same as $f\left(x\right)$ which is vertically stretched or compressed depending on value of $a$ .

Whereas for the function $f\left(ax\right)$ , using the concept of,

If $g\left(x\right)=f\left(bx\right)$

If $b>1$ it is known as Horizontal compression.

If $0<b<1$ it is known as Horizontal Stretch.

Therefore, the graph of the function $f\left(ax\right)$ is same as $f\left(x\right)$ which is horizontally stretched or compressed depending on the value of $a$ .

So, for $a\ne 1$ the graphs of functions $f\left(ax\right)$ and $af\left(x\right)$ are not same, therefore both of are not equivalent.

We conclude that,

Functions $f\left(ax\right)$ and $af\left(x\right)$ are equivalent if $a=1$ and not equivalent if $a\ne 1$ .