# Thread: Absolute value function (SAT Math Question)

1. ## Absolute value function (SAT Math Question)

For f(x)=|3x-17|, what is one possible value of a for which f(a)<a?

2. ## Re: Absolute value function (SAT Math Question)

Can you solve \displaystyle \begin{align*} |3a-17| &< a \end{align*}?

3. ## Re: Absolute value function (SAT Math Question)

Or look at the graph.

4. ## Re: Absolute value function (SAT Math Question)

We tried that and couldn't figure it out. I think this is what we did:

$|3a-17|

$-a<3a-17

$-a+17<3a

$\frac{-3+17}3

Not sure what to do from here

5. ## Re: Absolute value function (SAT Math Question)

That narrows it down, but not sure if you can get a good enough answer from that on the test (if she can use a calc on it)

6. ## Re: Absolute value function (SAT Math Question)

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").

7. ## Re: Absolute value function (SAT Math Question)

That narrows it down, but not sure if you can get a good enough answer from that on the test (if she can use a calc on it)
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.

8. ## Re: Absolute value function (SAT Math Question)

$|3a-17|

$-a<3a-17 At this point split the expression into two parts.

$-a<3a-17~~\&~~3a-17 solve each and intersect the solutions.

9. ## Re: Absolute value function (SAT Math Question)

Originally Posted by emakarov
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").
-a < 3a - 17
-a+17 < 3a
17 < 4a
17/4 < a

3a - 17 < a
2a < 17
a < 17/2

I'll put them back together?
17/4 < a < 17/2
matches the back of the book.
TY!

10. ## Re: Absolute value function (SAT Math Question)

Originally Posted by Plato
$|3a-17|

$-a<3a-17 At this point split the expression into two parts.

$-a<3a-17~~\&~~3a-17 solve each and intersect the solutions.
Thanks Plato. I finished the problem in another reply.