Order of transformations to inverse trig function For graphing $y=\cos^{-1}(1-2x)=
\cos^{-1} \left[-2\left(x-\dfrac{1}{2} \right) \right]$
I got this correct by finding the coordinates of the key points but when I trying plotting this just by applying transformations to the basic graph of arccos, I don't get it correct?
For this, we need to translate, then dilate and reflect, right?
I get the correct graph by doing the opposite order, but I'm pretty sure this order is incorrect..
Thanks
 A: If one knows what the graph of $y=\cos^{-1}x$ looks like, one could work it as follows. Afterwards, I will show how to proceed when one does not know what the graph of $y=\cos^{-1}x$ looks like.


*

*Begin with the graph of $y=\cos^{-1}x$



*To graph $y=\cos^{-1}(2x)=\cos^{-1}\left(\dfrac{x}{1/2}\right)$ shrink the graph by a factor of $\frac{1}{2}$ horizontally.





*To graph $y=\cos^{-1}\left(-\dfrac{x}{1/2}\right)$ reflect the previous graph about the $y$-axis.





*To graph $y=\cos^{-1}\left(-\dfrac{(x-\frac{1}{2})}{1/2}\right)$, shift the previous graph one-half unit to the right.



Even if one does not know what the graph of $y=\cos^{-1}x$ looks like one can still use transformations to figure it out so long as one know that reflecting a graph $f$ about the line $y=x$ produces the graph of the inverse and that for inverse functions the portion of the graph being reflected must satisfy the horizontal line test: no horizontal line can cross the graph twice. For $f(x)=\cos x$ that is the portion of the graph on the interval $[0.\pi]$.
In that case one would proceed as follows: 
If $y=\cos^{-1}(1-2x)$ then $0\le x\le1$ and $0\le y\le\pi$. Solving the equation for $x$ gives
$$ x=\frac{1}{2}(1-\cos y) \text{ for }0\le y\le\pi$$
You may instead graph
$$ y=\frac{1}{2}(1-\cos x) \text{ for }0\le x\le\pi$$
and reflect it in the line $y=x$ to produce the graph.
So your transformations should be made on this second graph.


*

*Graph $y=\cos x$ on the interval $[0,\pi]$.





*Reflect the graph in the $x-axis$ to obtain the graph of $y=-\cos x$.





*Adding $1$ shifts the graph up one unit to give the graph of $y=1-\cos x$.





*Multiplying by $\frac{1}{2}$ compresses the graph by half vertically to give the graph of $y=\frac{1}{2}(1-\cos x)$



*Reflect this graph in the line $y=x$ to obtain the graph of $y=\cos^{-1}(1-2x)$.



