How do you find $(g o h) (- 4)$ $(goh)(-4)$ given $g (a) = 2 a - 1$ $g(a)=2a-1$ and $h (a) = 3 a - 3$ $h(a)=3a-3$ ?

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How do you find $(g o h) (- 4)$ $(goh)(-4)$ given $g (a) = 2 a - 1$ $g(a)=2a-1$ and $h (a) = 3 a - 3$ $h(a)=3a-3$ ?

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Answer 1 · 2017-04-01T20:14:30

$(g \circ h) (- 4) = - 31$ $(g@h)(-4)=-31$

Explanation:

Note $(g \circ h) = g (h (x))$ $(g@h)=g(h(x))$

To find g(h(a)) substitute h(a) into g(a)

$g (h (a)) = g (3 a - 3) = 2 (3 a - 3) - 1$ $g(h(a))=g(color(red)(3a-3))=2(color(red)(3a-3))-1$

$= 6 a - 6 - 1$ $=6a-6-1$

$= 6 a - 7$ $=6a-7$

$\Rightarrow (g \circ h) (- 4)$ $rArr(g@h)(-4)$

$= (6 \times - 4) - 7 = - 31$ $=(6xx-4)-7=-31$

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How do you find $(g o h) (- 4)$ $(goh)(-4)$ given $g (a) = 2 a - 1$ $g(a)=2a-1$ and $h (a) = 3 a - 3$ $h(a)=3a-3$ ?

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How do you find $(g o h) (- 4)$ $(goh)(-4)$ given $g (a) = 2 a - 1$ $g(a)=2a-1$ and $h (a) = 3 a - 3$ $h(a)=3a-3$ ?