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Express the function in the form $ f \circ g \circ h $.

$ S(t) = \sin^2 (\cos t) $

Let $h(t)=\cos t, g(t)=\sin t,$ and $f(t)=t^{2}$. Then\[(f \circ g \circ h)(t)=f(g(h(t)))=f(g(\cos t))=f(\sin (\cos t))=[\sin (\cos t)]^{2}=\sin ^{2}(\cos t)=S(t)\].

02:24

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 3

New Functions from Old Functions

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

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all right. Here we see ass of tea and we want to break that down into a composition f of G of h. Another way to write that would be with parentheses, f of G of h, of tea, and we can see from that that the innermost function is H. Now, remember, when you're looking at acid T and you see the squared symbol right here, what that really means when we do that with trick functions is it means you're going to take the sine function and you're going to square it. So that's the order of those functions there. So now going back to h of tea. That's the innermost function. So if we look at this, the innermost part is the co sign of tea. So let's let h of t equal the co sign of tea Now the next layer would be G. H is inside g. So what's the next layer? We see the next layer, we see a sign. So if we let GFT equal the sine of T then noticed that if you were to put H inside G, you would have co sign inside sign and you would have signed of co sign T. That's what we want. Finally, we want the outermost Let your f So the outermost function is the squaring function. So let's let f of t equal t squared so that if we were to find f of g of h of tea, we would be squaring what we just had. And that's what we wanted. That's what s empty is

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