How am I to find a range of a function?

[Deleted member 4993]

what if I had y = |x-2| + 3 for -2≤x≤4? Should I do this? x-2 = 0, x = 2 so x = 5?

Again - first sketch the function. After sketching it 'correctly' - things would be clearer.

 

[Dr.Peterson]

Nevermind. @Dr.Peterson what if I had y = |x-2| + 3 for -2≤x≤4? Should I do this? x-2 = 0, x = 2 so x = 5?

You're right that x=2 is the turning point; but what are you saying about x=5??

 

[JeffM]

@Kulla_9289

You keep responding in very brief replies, and your English is not clear.

Do you understand that

(1) a question about a range will usually have an answer that is one or more intervals rather than points.

(2) a question about a function that includes one or more absolute values can be decomposed into a question about a function that includes no absolute values but is defined piecewise.

(3) for a function that is defined piecewise, the range of the entire function is the union of the ranges of the pieces.

(4) for a function that is defined on one or more closed intervals, determining the range of the function requires testing the value of the function at each end point of each interval and at each extremum (if any) within each interval.

If you truly understand each of those points fully, what is the source of any further confusion about any problem that can be decomposed into a problem about the range of linear functions? There is nothing difficult at all about the range of linear functions.

 

[Harry_the_cat]

Again - first sketch the function. After sketching it 'correctly' - things would be clearer.

Did you do this for your original question? What did you get? Graphing, in my opinion, is the easiest way to do both of these problems.
Graph y=x-2, then graph y=|x-2|, then graph y=|x-2|+3.
Calculate the values of y when x=-2 and when x=4. Plot them on your graph. Now you should be able to see the range.

 

[Kulla_9289]

Crystal clear. Thank you.

 

[Deleted member 4993]

Crystal clear. Thank you.

What did you get for answer?

 

[Harry_the_cat]

Crystal clear. Thank you.

Yes please share your answer.

 

[Kulla_9289]

3≤f(x)≤7 for f(x) = |x-2|+3.

 

[JeffM]

3≤f(x)≤7 for f(x) = |x-2|+3.

What you have literally written makes no sense.

[math]x = 100 \implies x - 2 = 98 > 0 \implies |100 - 2| = 98.\\ \therefore \ f(x) = |x - 2| + 3 \implies f(100) = 98 + 3 = 101 > 7. [/math]
However,

[math] \text {Given } f(x) = |x - 2| + 3 \text { for } - 2 \le x \le 4.\\ x - 2 = 0 \implies - 2 \le x \le 4 \text { and } f(x) = |0| + 3 = 3 \implies \\ f(x) \ge 3 \text { if } - 2 \le x \le 4.\\ - 2 - 2 = - 4 < 0 \implies |-2-2| = -(-4) = 4 \implies f(x) \le 4 + 3 = 7 \text { if } - 2 \le x < 2.\\ 4 - 2 = 2 > 0 \implies |4 - 2| = 2 \implies f(x) = 2 + 3 = 5 < 7.\\ \therefore 3 \le f(x) \le f(7). [/math]

 

[Harry_the_cat]

What you have literally written makes no sense.

[math] \therefore 3 \le f(x) \le f(7). [/math]

You mean 7 not f(7).
Kulla was asked for the range she/he found. She/he did that correctly.

 

[JeffM]

You mean 7 not f(7).
Kulla was asked for the range she/he found. She/he did that correctly.

Yes, you are correct. Thank you.

My point, however, is that the answer makes sense only with respect to the function’s being defined with respect to a limited domain.

That this thread is now at 71 posts is due to the utter failure of the original poster to bother with expressing things exactly. It is one unclear question after another. If you cannot formulate an exact question, you cannot complain about not getting exact answers.

 

[Kulla_9289]

I understand all these. But how would you write this as a piecewise function: f(x)=|x+2|+|x-2|
I had two piecewise functions.

 

[blamocur]

I understand all these. But how would you write this as a piecewise function: f(x)=|x+2|+|x-2|
I had two piecewise functions.

1. Figure out the points at which the function's simple (i.e., without absolute values) representation changes.
2. Figure out the formula for each resulting segment.
E.g. the representation of |x+2| changes at x=-2, and that of |x-2| at x=2. Thus we end up with 3 segments: [imath][-\infty,-2], [-2,2], [2,\infty][/imath]. You can now figure out how to express the function on each of these segments without using absolute values.

 

[Kulla_9289]

f(x) = {x+2 x≥-2 and -x-2 x<-2} and
f(x) = {x-2 x≥2 and -x+2 x<2}. Then?

 

[blamocur]

f(x) = {x+2 x≥-2 and -x-2 x<-2} and
f(x) = {x-2 x≥2 and -x+2 x<2}. Then?

g(x) = {x+2 x≥-2 and -x-2 x<-2} and
h(x) = {x-2 x≥2 and -x+2 x<2}. Then:

g(x) + h(x) = ....

 

[Kulla_9289]

g(x) + h(x) = { 0 -2≤x<-2 and 0 2≤x<2}

 

[blamocur]

g(x) + h(x) = { 0 -2≤x<-2 and 0 2≤x<2}

I don't understand this notation, nor does it look close to the correct answer. Here are 3 questions which need to be answered:

1. How many segments will there be in the definition of g+h ?
2. What are those segments?
3. What are the expressions for g+h on each of those segments?

 

[Kulla_9289]

I don't understand this notation, nor does it look close to the correct answer.

I added them. What is the definition of g + h?

 

[blamocur]

I added them. What is the definition of g + h?

"g+h" is a just shorthand for g(x)+h(x).

 

[Steven G]

1st of all, no sum adds to 0 so that is not correct.
You need to understand that for different intervals the summands will be different. You need to decide what intervals have g(x) and h(x) not changing, then and only then can you add them. Give it a try.