I need a solution or explanation for (c). I get stuck in this problem.

Subhotosh Khan said:
According to your textbook/class-notes:

What is the definition of "fg"?

Please share your work for parts 'a' & 'b'
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The definition of the fg (x) means a composite function and g acts on ( x ) first, then f conducts on the results. 03022B7F-E679-4DE3-A0D1-3B1FED1EDBC9.jpegThis is my solution for parts a and b. The results are checked out with the textbook‘s answers.AB874F60-63EE-448C-8FA7-186CD8CAF042.jpeg When I reached to part c to solve,I am not able to figure out this and didn't get the correct answer.
 
Eric Thet said:
The definition of the fg (x) means a composite function and g acts on ( x ) first, then f conducts on the results. View attachment 31033This is my solution for parts a and b. The results are checked out with the textbook‘s answers.View attachment 31034 When I reached to part c to solve,I am not able to figure out this and didn't get the correct answer.
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Thanks for showing what you've done, though it might have been helpful not to erase the work for (c), in case you were closer than you thought.

The domain will be the set on which the radicand is non-negative (and exists). So solve the inequality [math]\frac{x-5}{2x+1}\ge 0[/math]
Do you know any methods for doing that?
 
pka said:
To Eric Thet, As written it is domain & range of [imath]\Large fg[/imath] not [imath]\Large f\large \circ g[/imath]
I.e, product as opposed to composition.
Which one is it?

[imath][/imath][imath][/imath]
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According to his work,

fg = [imath]\Large f\large \circ g[/imath] ....................... hence it is the composition.
 
Dr.Peterson said:
Thanks for showing what you've done, though it might have been helpful not to erase the work for (c), in case you were closer than you thought.

The domain will be the set on which the radicand is non-negative (and exists). So solve the inequality [math]\frac{x-5}{2x+1}\ge 0[/math]
Do you know any methods for doing that?
Click to expand...
Even though we don't gain the answer to this by solving it.
 
Last edited:
He said it w
Eric Thet said:
Even though we don't gain the answer to this by solving it.
Click to expand...
I must admit I find Dr. Peterson’s answer here to be a bit obscure.

I would say [imath]2x + 1 > 0.[/imath]

But therefore

[math]2x + 1 > 0 \implies \dfrac{1}{2x + 1} > 0 \text { and}\\ 2x + 1 > 0 \implies x > - \dfrac{1}{2} \implies x + 5 > \dfrac{9}{2} > 0 \ge 0.\\ \therefore \dfrac{x + 5}{2x + 1} \ge 0.[/math]
Note that I am NOT saying his answer is wrong. I’d just prefer to state the constraint differently, namely as [imath]2x + 1 > 0 \implies x > - \dfrac{1}{2}.[/imath]
 
JeffM said:
He said it w

I must admit I find Dr. Peterson’s answer here to be a bit obscure.

I would say [imath]2x + 1 > 0.[/imath]

But therefore

[math]2x + 1 > 0 \implies \dfrac{1}{2x + 1} > 0 \text { and}\\ 2x + 1 > 0 \implies x > - \dfrac{1}{2} \implies x + 5 > \dfrac{9}{2} > 0 \ge 0.\\ \therefore \dfrac{x + 5}{2x + 1} \ge 0.[/math]
Note that I am NOT saying his answer is wrong. I’d just prefer to state the constraint differently, namely as [imath]2x + 1 > 0 \implies x > - \dfrac{1}{2}.[/imath]
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Are we answering different questions?

The question, as I understand it, is to find the domain of [imath]f\circ g(x) = \sqrt{\frac{x-5}{2x+1}}[/imath].

You are claiming, I think, that the domain is [imath]x>-\frac{1}{2}[/imath].

But, for example, if [imath]x=0[/imath], then [imath]f\circ g(0) = \sqrt{\frac{0-5}{2(0)+1}}=\sqrt{-5}[/imath], which is undefined.

And, on the other hand, if [imath]x=-1[/imath], then [imath]f\circ g(-1) = \sqrt{\frac{-1-5}{2(-1)+1}}=\sqrt{6}[/imath], which is defined.
 
Dr.Peterson said:
Are we answering different questions?

The question, as I understand it, is to find the domain of [imath]f\circ g(x) = \sqrt{\frac{x-5}{2x+1}}[/imath].

You are claiming, I think, that the domain is [imath]x>-\frac{1}{2}[/imath].

But, for example, if [imath]x=0[/imath], then [imath]f\circ g(0) = \sqrt{\frac{0-5}{2(0)+1}}=\sqrt{-5}[/imath], which is undefined.

And, on the other hand, if [imath]x=-1[/imath], then [imath]f\circ g(-1) = \sqrt{\frac{-1-5}{2(-1)+1}}=\sqrt{6}[/imath], which is defined.
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You are right as usual. I had x + 5 in the numerator.
 
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