No, jumbogala, you were right here and, unfortunately, berkeman was wrong (I'll never let him live it down!). "Piecewise continuous" means "continuous in pieces"- that is that it is continuous everywhere except possibly at a finite number of point, separating the continuous "pieces".
And, berkeman, he does NOT "want to use a step funtion for x≤8", he wants to use a step function for the entire function. That is, write a formula, including a step function so that the entire function is given by that step function.
Let's see you suggested f(x)= 1-(-8)(h(x-8)) + (-x-1)(h(x-8)). Let's check to see if that is correct. h(x) is 0 if x< 0 so h(x-8) is 0 if x is less than 8. If x< 8 then f(x)= 1-(-8)(0)+ (-x-1)(0)= 1. No, that isn't what you want- you want -8. Also h(x-8) is equal to 1 if x is larger than or equal to 8. If x is larger than or equal to 8, f(x)= 1- (-8)(1)+(-x-1)(1)= 1+8- x-1= 8- x. No, that isn't it either.
Think about this: If f(x)= u(x) for x< a and f(x)= v(x) for x\ge a, and we want to write it as f(x)= p(x)h(x-a)+ q(x), what must p and q equal? Well, if x< a, h(x-a)= 0 so f(x)= p(x). Obviously, we must have p(x)= u(x). If x\ge a, h(x-a)= 1 so f(x)= u(x)+ q(x)(1)= v(x) so q(x)= v(x)- u(x).
In this problem u(x)= -8 and v(x)= -x-1
