Question #6be76

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Question #6be76

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Answer 1 · 2016-10-20T21:59:13

(i) $f \circ h (x) = sin (3 x^{2} - 1)$ $f@h(x)=sin(3x^2-1)$
(ii) $h \circ g (x) = 3 {cos}^{2} (x) - 1$ $h@g(x)=3cos^2(x)-1$
(iii) $g \circ f (x) = cos (sin x)$ $g@f(x)=cos(sinx)$
$g o f (0) = 1$ $gof(0)=1$

Explanation:

(i) $f \circ h (x) = f (h (x)) = sin (h (x)) = sin (3 x^{2} - 1)$ $f@h(x)=f(h(x))=sin(h(x))=sin(3x^2-1)$

(ii) $h \circ g (x) = h (g (x)) = 3 {(g (x))}^{2} - 1 = 3 {cos}^{2} (x) - 1$ $h@g(x)=h(g(x))=3(g(x))^2-1=3cos^2(x)-1$

(iii) $g \circ f (x) = g (f (x)) = cos (f (x)) = cos (sin x)$ $g@f(x)=g(f(x))=cos(f(x))=cos(sinx)$

(iv) $g \circ f (0) = g (f (0))$ $g@f(0)=g(f(0))$
Using the computed composition in (iii):

$g o f (0) = cos (sin 0) = cos (0) = 1$ $gof(0)=cos(sin0)=cos(0)=1$

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Question #6be76

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