transforming the graph of a inverse trigonometric function

Question

I don't understand how to sketch properly, for example $f(x) = \arccos (2x+1)$, what I find particularly confusing is how the order of transformations should be understood.

The composite function can be decomposed as $$x \xrightarrow{\cdot 2} 2x \xrightarrow{+ 1} 2x + 1 \xrightarrow{\arccos} \arccos (2x+1).$$

Graphically: given the graph of $\arccos$, apply

• 1st: horizontal stretch by the scale factor $\sigma=\frac{1}{2}$ (so the domain is half in size)
• 2nd: horizontal translation to the left by $1$ unit

If applied in this order the graph is wrong, see graph of $f(x)$. To make it consistent with it seems the the order should be reversed, 1st the translation, then the stretch. I don't understand why. Is it related to some co- / contra-variance features that I missed?

Answer 1

1. Start with the graph of $$y=f(x).$$ Applying a horizontal scale factor of $\frac12$ gives $$y=f(2x):=g(x).$$ Translating the graph by $1$ unit to the left gives $$y=g(x+1)=f(2x+2).$$

Your above procedure resulted in the graph of $\arccos(2x+2).$ On the other hand, the procedure below gives the desired $\arccos(2x+1).$

2. Start with the graph of $$y=f(x).$$ Translating the graph by $1$ unit to the left gives $$y=f(x+1):=g(x).$$ Applying a horizontal scale factor of $\frac12$ gives $$y=g(2x):=f(2x+1).$$

Here's a related discussion: How does one explain that transformations 'inside' a function operate in the opposite direction than intuition suggests?.