Function: If (fog)(x)=(sin squar x) ^ 2 and (gof)(x)=|sinx|, find f(x) and g(x)

[minerwamin]

If (fog)(x)=(sin squar x) ^ 2 and (gof)(x)=|sinx|, find f(x) and g(x)

 

[stapel]

If (fog)(x)=(sin squar x) ^ 2 and (gof)(x)=|sinx|, find f(x) and g(x)

What are your thoughts? What have you tried? How far have you gotten? ("Read Before Posting")

When you reply, please include a clarification of the expression for $(f \circ g)(x)$. In particular, it looks as though you are squaring the sine twice: $\left(\sin^2(x)\right)^2$. Is this correct?

Thank you!

 

[Dr.Peterson]

If (fog)(x)=(sin squar x) ^ 2 and (gof)(x)=|sinx|, find f(x) and g(x)

I suspect you may have written "squar" meaning "square root", which is commonly abbreviated as "sqrt":

(fog)(x)=(sin sqrt(x))^2​

(gof)(x)=|sinx|​

$(f\circ g)(x)=(\sin(\sqrt{x}))^2$​

$(g\circ f)(x)=|\sin(x)|$​

I see three things being done in the first composite function (square root, sine, square), so it seems likely that f does one of those, and g does the other two, or vice versa. How can you split up those actions so that swapping the order would result in the second composite function? (Think about how squares and square roots interact.)

 

[skeeter]

note …

$\sqrt{\sin^2{x}} = |\sin{x}|$

 

[Steven G]

f(x)=x and g(x) = (sin²x)² or g(x)=x and h(x) = (sin²x)².
This never fails.

 

[Dr.Peterson]

f(x)=x and g(x) = (sin²x)² or g(x)=x and h(x) = (sin²x)².
This never fails.

What problem are you solving?

"Decomposition" problems like $(f\circ g)(x)=(\sin(\sqrt{x}))^2$ alone can be tricky to pose so as to elicit a suitable answer, because of the sort of thinking you imply; but this one is considerably less ambiguous (once stated clearly).

 

[Steven G]

What problem are you solving?

The (incorrect) problem that the student posted.

 

[Dr.Peterson]

The (incorrect) problem that the student posted.

Really? How do you get

If (fog)(x)=(sin squar x) ^ 2 and (gof)(x)=|sinx|, find f(x) and g(x)

$(f\circ g)(x)=(\sin(x^2))^2$ and $(g\circ f)(x)=|\sin(x)|$ from your two functions?

f(x)=x and g(x) = (sin²x)² or g(x)=x and h(x) = (sin²x)².

Your $g(x)=\sin^4(x)$; and what is h(x)? I expected you to make some sort of correction.

 

[Steven G]

OK, I misread the question. I was just basically saying that whenever you have h(x) = (fog)(x) you can define f(x)=x and g(x) = h(x) and never be wrong.

 

[Dr.Peterson]

OK, I misread the question. I was just basically saying that whenever you have h(x) = (fog)(x) you can define f(x)=x and g(x) = h(x) and never be wrong.

Yes, that's what I referred to when I said

"Decomposition" problems like $(f\circ g)(x)=(\sin(\sqrt{x}))^2$ alone can be tricky to pose so as to elicit a suitable answer, because of the sort of thinking you imply; but this one is considerably less ambiguous (once stated clearly).

But, of course, this isn't that sort of problem, typo or not, as it gives both fog and gof, not just one of them.