Function Into Function 3

[mathdad]

See attachment.

 

[ksdhart2]

Well, I mean, your answer is correct, but I'd get rid of the "Let \(x = x \sin(x)\) line." That's a confusing, and generally untrue, statement, that's never actually used in the problem so having it there just makes everything worse and harder to understand. If you wanted to go the substitution route you could say:

Let \(\displaystyle y = h(x) = x \sin(x)\)

Such that the problem boils down to:

\(\displaystyle h(h(x)) = h(y) = y \sin(y)\)

And then by back-substituting \(y\) you get:

\(\displaystyle h(h(x)) = y \sin(y) = x \sin(x) \sin\big(x \sin(x) \big)\)

 

[mathdad]

Well, I mean, your answer is correct, but I'd get rid of the "Let \(x = x \sin(x)\) line." That's a confusing, and generally untrue, statement, that's never actually used in the problem so having it there just makes everything worse and harder to understand. If you wanted to go the substitution route you could say:

Let \(\displaystyle y = h(x) = x \sin(x)\)

Such that the problem boils down to:

\(\displaystyle h(h(x)) = h(y) = y \sin(y)\)

And then by back-substituting \(y\) you get:

\(\displaystyle h(h(x)) = y \sin(y) = x \sin(x) \sin\big(x \sin(x) \big)\)

Ok.