How am I to find a range of a function?

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Kulla_9289

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Say that I am given such: f(x) = 5-|2x| for -3≤x≤4. How do I find the range?
 
Kulla_9289 said:
Say that I am given such: f(x) = 5-|2x| for -3≤x≤4. How do I find the range?
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If I were to solve the problem - I'll draw an approximate sketch of the function within the given domain.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
I don't know how to sketch this. I prefer algebraic method. I know how to sketch if it were f(x) = |5-2x|
 
Kulla_9289 said:
Say that I am given such: f(x) = 5-|2x| for -3≤x≤4. How do I find the range?
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x = -3 provides f(x) = y = -1

x = -2 provides f(x) = y = 1 .......................x = -2.5 provides f(x) = y = 0

x = -1 provides f(x) = y =3

x = 0 provides f(x) = y = 5

x = 1 provides f(x) = y =3

x = 2 provides f(x) = y = 1 .......................x = 2.5 provides f(x) = y = 0

x = 3 provides f(x) = y = -1 .......................x = 2.5 provides f(x) = y = 0

Continue...

Now plot these points on a piece of paper with x & y axes and origin - and analyze.......
 
Even I came up with this rudimentary method. I do not know the more efficient method, that's why I asked. I know modulus. But I have seen in a book that they have equated it to zero which I don't understand why
 
Kulla_9289 said:
I do not know the more efficient method, that's why I asked. I know modulus. But I have seen in a book that they have equated it to zero which I don't understand why
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If you're asking about a method you've seen, why don't you show it to us, so we can help you understand that, rather than guess?

Show us an image of what the book says, or a link if it's available online.
 
Here
 

Attachments

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    Screenshot_20221010_204427.jpg
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There are almost 10 pages in your attachments. Do you have a question about a specific statement or formula?
Also, have you tried the approach from post #6? If you have, what have you gotten?
 
Why in the world would you suggest that equating a function to zero is a method for finding a function‘s range? That makes no sense.

There is no general algebraic method for determining a function’s range. Differential calculus works for functions that are differentiable everywhere.
 
Kulla_9289 said:
Here
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What they're doing on page 129, the image you attached separately, is just graphing the function and looking at the range. That's just what was suggested, and is often the most efficient method.

Everywhere I find the word "range" in the pdf, they are doing the exact same thing.

Kulla_9289 said:
Say that I am given such: f(x) = 5-|2x| for -3≤x≤4. How do I find the range?
Click to expand...
Kulla_9289 said:
I would equal it to zero. But I am not sure why would we do equal it to zero
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Kulla_9289 said:
Even I came up with this rudimentary method. I do not know the more efficient method, that's why I asked. I know modulus. But I have seen in a book that they have equated it to zero which I don't understand why
Click to expand...
When they set y to 0, it is to find the x-intercepts, if any, not to find the range. So you are misunderstanding what you read.

Here is how I would approach your example, f(x) = 5-|2x| for -3≤x≤4:

First, observe that the function changes its behavior when we take the absolute value of 0, namely at x=0. So I would first plot (x,y) for x=0, -3, 4, which are end points of parts of the domain. Then I would observe that within each region, the graph will be linear, so I would join the plotted points with line segments.

At this point, I have the graph, and can determine the range, which will be from the lowest of the three points to the highest.

I don't think there's a quicker way, unless I just type the equation into Desmos.
 
Kulla_9289 said:
@blamocur check the screenshot. @JeffM I know. I can differentiate x² + 6x + 4 for x≥0, but I don't know for this one.
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You can always treat a function with absolute values as a piecewise function. If all the pieces are differentiable, you can differentiate within the intervals that are defined piecewise and calculate values at extreme points and at all boundary points.

Of course, in your example, breaking the function down into a piecewise function results in two linear functions, so compute the values at the boundaries.
 
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