A basic function question - with a strange absolute value placement

AI Thread Summary
The discussion centers on the function f(x) defined in three segments: x+9 for x < -3, -2x for |x| ≤ 3, and -6 for x > 32. The placement of the absolute value in the middle segment is clarified to be significant, as |x| ≤ 3 translates to the range -3 ≤ x ≤ 3. This means the function is defined for all x within that range, contrary to the initial assumption that it might be undefined between -3 and 0. The clarification emphasizes that the absolute value notation is not a distraction but an essential part of the function's definition. Understanding this equivalence resolves the confusion regarding the function's domain.
latefreight
Messages
2
Reaction score
0

Homework Statement



If f(x) is equal to...

x+9 if x<-3
-2x if |x|\leq 3
-6 if x > 32. The attempt at a solution

The first and third "segments" of when the function is defined as being x+9 and -6 are pretty straightforward to me, however I am unaware of the significance of the placement of the absolute value of x in the middle if statement of when the function is equal to -2x. Is it significant at all? Does this imply this function is undefined when -3 < x < 0? Or is it, as my initial hunch was, really just some type of a distraction?
 
Physics news on Phys.org
|x| <= 3 is equivalent to -3 <= x <= 3, so your hunch that this is a distraction is wrong.
 
<br /> 0 \le \left| x \right| \le a \; \Leftrightarrow \; -a \le x \le a<br />
 
Thank you. I didn't choose my word "distraction" cautiously enough. Since the statements you both posted are equivalent - that was indeed what I needed to know.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Replies
18
Views
2K
Replies
26
Views
3K
Replies
10
Views
2K
Replies
10
Views
8K
Back
Top