how do you sketch h(root x) for question 5?
Let's break it down. For x<0, the function will have no value, since the square root of a negative number has no real value in the real domain. So we can disregard that completely.
Now, the current function has roots at x=0, x=2, x=4, x=6 etc. This will remain true, however, it is now the square root of x which must equal these values! So:
So the intercepts will be in these locations. Essentially, what is happening here is that the triangular wave is being stretched out more and more with larger values of x. The further up you go, the further apart the intercepts. This is easiest to see if you look at the graph over a large domain.
To get a clue as to the shape itself, consider x=0 to x=1. In this region, we can model the behaviour of the curve with the function:
Since in the original, the function in this domain was equivalent to y=x. What this (and probably plotting a few points would help if you are unsure) tells us is that while the triangular shape is still present, it is now slightly curved. The square root sign, beyond stretching out the waveform, is also adding the curved shape that we see in the graph of y equals the square root of x.
The graph looks like this. This is an extremely difficult question, although I suspect there might be some 4 Unit processes that eliminate the more intuitive approach I have taken. I am unsure exactly what you've covered, so I stuck with intuition