Functions

Did you mean
1651234047982.png

I would do the evaluation of f(g(f(g(1/2)))) step by step, rather than trying to write a single expression.

What is g(1/2)? It's -2(1/2) = -1, because 1/2 > 0.

What is f(-1)?

And so on.
 
I would approach this one step at a time:

[math]f \left(\frac{1}{2}\right) = ?[/math][math]g\circ f \left(\frac{1}{2}\right) = ?[/math],
[math]f \circ g\circ f \left(\frac{1}{2}\right) = ?[/math],
[math]g \circ f \circ g\circ f \left(\frac{1}{2}\right) = ?[/math],
 
Dr.Peterson said:
Did you mean
View attachment 32468

I would do the evaluation of f(g(f(g(1/2)))) step by step, rather than trying to write a single expression.

What is g(1/2)? It's -2(1/2) = -1, because 1/2 > 0.

What is f(-1)?

And so on.
Click to expand...
I didn't even notice that. However, I checked, it's the same as I have written it. Probably a mistake.
Would you start evaluating from left or right side?
 
blamocur said:
I would approach this one step at a time:

[math]f \left(\frac{1}{2}\right) = ?[/math][math]g\circ f \left(\frac{1}{2}\right) = ?[/math],
[math]f \circ g\circ f \left(\frac{1}{2}\right) = ?[/math],
[math]g \circ f \circ g\circ f \left(\frac{1}{2}\right) = ?[/math],
Click to expand...
Just realized that I was echoing @Dr.Peterson, but at least my typesetting is fancier :)
 
Loki123 said:
I didn't even notice that. However, I checked, it's the same as I have written it. Probably a mistake.
Would you start evaluating from left or right side?
Click to expand...
Evaluate it as the notation requires. Both of us (mister fancy and I;)) showed you where to start.
 
Loki123 said:
Would you start evaluating from left or right side?
Click to expand...
I hadn't noticed until now that @blamocur showed how to answer the wrong question! What he said would be correct if you had wanted [imath]g \circ f \circ g\circ f \left(\frac{1}{2}\right)[/imath]; but you want [imath]f \circ g \circ f\circ g \left(\frac{1}{2}\right)[/imath].
 
I need reminding of the definitions of the functions. [Wish the TXT complier were better}
[imath]f(x)=-2\text{ if }x\le -1,\;|x|\text{ if }, -1< x<1,\;x+1\text{ if }1\le x[/imath]
[imath]g(x)=\;0\text{ if }x\le0, -2^x\text{ if }0<x[/imath]
Are you clear on the fact that [imath]g\left(\tfrac{1}{2}\right)=-\sqrt{2}~?[/imath]
If you are clear on that then [imath]f\circ g\left(\tfrac{1}{2}\right)=-2[/imath]
Can you explain this for any [imath]n\in\mathbb{N}^+[/imath]. [imath]\underbrace {f \circ f \circ \cdots f \circ f}_n\left( { - \sqrt 2 } \right) = -2~~?[/imath]

[imath][/imath]
 
Dr.Peterson said:
I hadn't noticed until now that @blamocur showed how to answer the wrong question! What he said would be correct if you had wanted [imath]g \circ f \circ g\circ f \left(\frac{1}{2}\right)[/imath]; but you want [imath]f \circ g \circ f\circ g \left(\frac{1}{2}\right)[/imath].
Click to expand...
I guess I did not echo you correctly :( -- thanks for the correction.
 
Looks good me.
My only concern is that you never answered the question in my opinion!
You should have somewhere that fogofog(1/2) = 0. In my opinion this is important. I would definitely take off some credit on an exam for failing to state the answer.
 
Last edited:
Steven G said:
Looks good me.
My only concern is that you never answered the question in my opinion!
You should have somewhere that fogofog(1/2) = 0. In my opinion this is important. I would definitely take off some credit on an exam for failing to state the answer.
Click to expand...
okay i will keep that in mind!
 
You must log in or register to reply here.
Top