Sketching graphs abs value functions how do I go forward with sketching the graphs of the following two functions?
i) $y(t)=|2+t^3|$
ii) $f(x)=4x+|4x-1|$
thanks in advance!
 A: Hints: 
Trick is to see when $| P(x) |$ or $|P(t)|$changes its value. That is $|x|$ is defined to be equal to $x$ when $x \ge 0$ and $(-x)$ when  $x \lt 0$. So check at which point/s the expressions within the absolute value brackets changes its sign. Now demarcate the two areas split by this point. On either side of this point on the real line you will have equations of two different straight lines. Should be easy from there. 
A: For functions like the second one, where everything is linear functions of x, I'd actually just break it up algebraically.
When $x \ge 1/4$, $4x-1 \ge 0$ and so $|4x-1|=4x-1$. So in this range, $f(x) = 4x+(4x-1) = 8x-1$.
When $x \lt 1/4$, $4x-1 \lt 0$ and so $|4x-1|=-(4x-1)=1-4x$. So in this range, $f(x) = 4x+(1-4x)=1$.
In other words, to the left of $x=1/4$, it's a flat line at $y=1$, and to the right it's the line $y=8x-1$.
Sketching the graphs of $y=4x$ and $y=|4x-1|$ on the same graph should also hint at this behaviour, since to the left of the critical point they're moving in opposite directions so adding them will clearly cancel out in some fashion, while on the right they're parallel and increasing so adding them will give you something that increases even faster.
