How am I to find a range of a function?

@blamocur I am not sure how for the domain of the piecewise function. I think I need to combine them. So, f(x) = {5-2x 0≤x≤4 and 5+2x -3≤x<0}. Then?
 
Last edited:
The whole point of the piecewise approach to functions that contain the absolute function is this:

you break the domain up into pieces where the functions involving absolute values are replaced with functions that do not involve the absolute value function. With regard to [imath]-|2x|[/imath] we replace it with [imath]-(+2x) = - 2x[/imath]] where 2x is non-negative and with[imath]-(-2x) = +2x[/imath] where 2x is negative. We end up with [imath]5 - 2x[/imath] where x is non-negative and [imath]5 + 2x[/imath] where x is negative.

What did I say about using calculus? I said to compute values at extrema and boundaries. In the case of 5 - 2x and 5 + 2x, do they have extrema?
 
Steven G said:
Yes! Think what y values you 'd get if you plug in -3,-2,-1,0,1,2,3,4 for x. That will save ink and be much faster.
Click to expand...
Need to add x = -2.5 - where the function changes direction.
 
Kulla_9289 said:
Say that I am given such: [imath]\large{f(x) = 5-|2x| \text{ for } -3≤x≤4}[/imath]. How do I find the range?
Click to expand...
Kulla_9289, you are making far to much out of a simple pre-calculus algebra question.
The given function is not a piecewise function.
It is an even function on [imath][-3,3][/imath]: (i.e.) [imath]f(x)=f(-x)[/imath].
Do you see that [imath]f(-3)=f(3);~f(-2.5)=0=f(2.5),~\&~f(-1)=f(1)=3[/imath]
Look HERE at the graph. Use that graph to verify each of the above.
 
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 don't know how to sketch this. I prefer algebraic method. I know how to sketch if it were f(x) = |5-2x|
Click to expand...
Have you learnt about transformations of graphs?

Can you graph \(\displaystyle y=2x\)?

Can you graph \(\displaystyle y=|2x|\)?

Can you graph \(\displaystyle y=-|2x|\)?

Can you graph \(\displaystyle y=-|2x|+5\)?

You should now be able to work out the range for the domain you were given!
 
@blamocur how would I know which sign to put in my final answer?
@Dr.Peterson say that I have f(x)=x²+6x+4 for -5<x≤6. I differentiate the quadratic function to get 2x + 6. I equate it to zero to find its critical point. 2x + 6 = 0, x=-3. Is this in the range? Yes. So f(-3)= (-3)² + 6(-3) + 4, which gives -5. Then evaluate the endpoints, which give f(-5)=-1 and f(6)= 76. Absolute maximum: (6,76). Absolute minimum: (-3,-5). How would I which inequality sign to put in the final answer in terms of inequality? How would I write it? -5≤x≤76?
 
Harry_the_cat said:
Have you learnt about transformations of graphs?

Can you graph \(\displaystyle y=2x\)?

Can you graph \(\displaystyle y=|2x|\)?

Can you graph \(\displaystyle y=-|2x|\)?

Can you graph \(\displaystyle y=-|2x|+5\)?

You should now be able to work out the range for the domain you were given!
Click to expand...
No
 
Kulla_9289 said:
@blamocur how would I know which sign to put in my final answer?
Click to expand...
I don't know what you mean by this question. Have you found the ranges of each piece of the piecewise function? If you have can you post them?
 
Kulla_9289 said:
@Dr.Peterson say that I have f(x)=x²+6x+4 for -5<x≤6. I differentiate the quadratic function to get 2x + 6. I equate it to zero to find its critical point. 2x + 6 = 0, x=-3. Is this in the range? Yes. So f(-3)= (-3)² + 6(-3) + 4, which gives -5. Then evaluate the endpoints, which give f(-5)=-1 and f(6)= 76. Absolute maximum: (6,76). Absolute minimum: (-3,-5). How would I [know] which inequality sign to put in the final answer in terms of inequality? How would I write it? -5≤x≤76?
Click to expand...
How did you choose that inequality? Why are you not sure of it?
 
You must log in or register to reply here.
Top