For f(x)=|3x-17|, what is one possible value of a for which f(a)<a?
Reply to ProveIt:
We tried that and couldn't figure it out. I think this is what we did:
$\displaystyle |3a-17|<a$
$\displaystyle -a<3a-17<a$
$\displaystyle -a+17<3a<a+17$
$\displaystyle \frac{-3+17}3<a<\frac{a+17}3$
Not sure what to do from here
By definition, -a < 3a - 17 < a is two inequalities: -a < 3a - 17 and 3a - 17 < a. Solve both inequalities, take the intersection of the sets of their solutions (i.e., find the set of a's that are solutions to both inequalities) and pick any element of the intersection (since the question asks for "one possible value").
The graph can give you an idea for which x to try. Then you can pick an x and simply by calculation prove that it satisfies the inequality.
In the proofs of existence, you have to provide a witness (x in this case) and a proof that the witness satisfies the necessary property. You don't have to explain how you picked the witness unless it is a requirement of this particular problem/instructor/course.