According to your textbook/class-notes:
The definition of the fg (x) means a composite function and g acts on ( x ) first, then f conducts on the results. This is my solution for parts a and b. The results are checked out with the textbook‘s answers. When I reached to part c to solve,I am not able to figure out this and didn't get the correct answer.According to your textbook/class-notes:
What is the definition of "fg"?
Please share your work for parts 'a' & 'b'
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 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.
To Eric Thet, As written it is domain & range of [imath]\Large fg[/imath] not [imath]\Large f\large \circ g[/imath]
According to his work,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]
But what if it is a product and he read incorrectly??According to his work,
fg = [imath]\Large f\large \circ g[/imath] ....................... hence it is the composition.
Even though we don't gain the answer to this by solving it.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?
I don't know what you mean by this. The solution to this inequality is the domain of the composite function.Even though we don't gain the answer to this by solving it.
I must admit I find Dr. Peterson’s answer here to be a bit obscure.Even though we don't gain the answer to this by solving it.
Are we answering different questions?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]
You are right as usual. I had x + 5 in the numerator.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.
In that case, the domain would be [imath](-\infty,-5]\cup(-\frac{1}{2},\infty)[/imath].You are right as usual. I had x + 5 in the numerator.