1) a) For the following functions, find \( f \circ g \circ h \). \( f(x)=2 x-1, g(x)=x^{2}, h(x)=1-x \) b) Given the function \( F(x)=\sin ^{2}(x-3) \), find the functions \( f, g \) and \( h \) such that \( F=f \circ g \circ h \)

1) a) We are given:   f(x) = 2x – 1,   g(x) = x²,   h(x) = 1 – x. To find (f ∘ g ∘ h)(x), start with h(x), then apply g, and finally f. Step 1: Compute h(x):   h(x) = 1 – x. Step 2: Compute g(h(x)):   g(h(x)) = (1 – x)².   Expanding: (1 – x)² = 1 – 2x + x². Step 3: Compute f(g(h(x))):   f(g(h(x))) = f((1 – x)²) = 2·(1 – x)² – 1.   Substitute the expanded form:    = 2(1 – 2x + x²) – 1    = 2 – 4x + 2x² – 1    = 2x² – 4x + 1. Thus, (f ∘ g ∘ h)(x) = 2x² – 4x + 1. ------------------------------------------------------ 1) b) We are given:   F(x) = sin²(x – 3). A convenient choice is to rewrite F(x) in a way that fits the format f(g(h(x))). Notice that:   sin²(x – 3) = [sin(x – 3)]². One possible decomposition is: • Let h(x) = x – 3. • Let g(x) = sin x. • Let f(x) = x². Then, we have:   (g ∘ h)(x) = g(h(x)) = sin(x – 3),   and   (f ∘ g ∘ h)(x) = f(sin(x – 3)) = [sin(x – 3)]² = F(x). These choices satisfy the requirement, so the functions are:   h(x) = x – 3,   g(x) = sin x,   f(x) = x².