You're welcome. I'm glad I could help.

[lovemake1]

Homework Statement

Find an upper bound M for f(x) = abs ( x+2 / x-8 ) if abs(x-7) < 1/2

Homework Equations

The Attempt at a Solution

i first found set of x values using abs(x-7) < 1/2
which is 13/2 < x < 15/2.

Now, i believe i have to find other set of x values to compare to find upper bound for x.
But how ? please help.

 

[Mark44]

Homework Statement

Find an upper bound M for f(x) = abs ( x+2 / x-8 ) if abs(x-7) < 1/2

Homework Equations

The Attempt at a Solution

i first found set of x values using abs(x-7) < 1/2
which is 13/2 < x < 15/2.

Now, i believe i have to find other set of x values to compare to find upper bound for x.
But how ? please help.

You want to find an upper bound for f(x), not x. If 13/2 < x < 15/2, what are the possible values for |x + 2|/|x - 8|? Look at the numerator and denominator separately, and see what intervals you get for |x + 2| and |x - 8|.

 

[HallsofIvy]

You don't really need to find the x-limits. If -1/2< x- 7< 1/2, then, adding 9 to each part, 17/2< x+ 2< 19/2. Subtracting 1 from each part, -3/2< x- 8< -1/2.

To make a fraction as large as possible, make the numerator as large as possible and the denominator as small as possible.

 

[joshiemen]

You don't really need to find the x-limits. If -1/2< x- 7< 1/2, then, adding 9 to each part, 17/2< x+ 2< 19/2. Subtracting 1 from each part, -3/2< x- 8< -1/2.

To make a fraction as large as possible, make the numerator as large as possible and the denominator as small as possible.

Does that mean then

A) (x+2)<19/2 and (x-8)<-3/2, thus (x+2)/(x-8)<-19/3

i.e upper bound <-19/3?

or

B) |x+2|<19/2 and |x-8|<1/2 thus |x+2|/|x-8|<19

i.e upper bound <19 ?

 

[HallsofIvy]

If you think (A) is possible then (i) you did not read what I said (because you are making both numerator and denominator as large as possible) and (ii) you are not using common sense (because it makes no sense to say that a fraction of absolute values has a negative number as upper bound).

I said, "To make a fraction as large as possible, make the numerator as large as possible and the denominator as small as possible."

The numerator, |x+ 2|, can be no larger than 19/2. The denominator, |x- 8|, can be no smaller than 1/2 (-3/2< x- 8< -1/2 so 1/2< |x- 8|< 3/2).

\frac{|x+2|}{|x- 8|} can be no larger than \frac{19/2}{1/2}= 19.

 

[joshiemen]

If you think (A) is possible then (i) you did not read what I said (because you are making both numerator and denominator as large as possible) and you are not using common sense (because it makes no sense to say that a fraction of absolute values has a negative number as upper bound).

I said, "To make a fraction as large as possible, make the numerator as large as possible and the denominator as small as possible."

The numerator, |x+ 2|, can be no larger than 19/2. The denominator, |x- 8|, can be no smaller than 1/2 (-3/2< x- 8< -1/2 so 1/2< |x- 8|< 3/2).

\frac{|x+2|}{|x- 8|} can be no larger than \frac{19/2}{1/2}= 19.

HallsofIvy thanks for the reply, and appreciate the clarification, things are making sense now :)!