2 Answers

Flak DiNenno · Answer 1 · 2021-10-07T23:29:17+0000

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Work Shown:

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

$g(x) = x^2\\\\g(h(x)) = (h(x))^2 \ \text{replace every x with h(x)}\\\\g(h(x)) = (x-7)^2 \ \text{replace h(x) with x-7}\\\\(g \circ h)(x) = (x-7)^2$

Now plug in x = 5

$(g \circ h)(x) = (x-7)^2 \\\\(g \circ h)(5) = (5-7)^2 \\\\(g \circ h)(5) = (-2)^2\\\\ (g \circ h)(5) = 4$

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Alternative approach:

h(x) = x-7

h(5) = 5-7

h(5) = -2

So this means

g(h(5)) = g(-2)

and

g(x) = x^2

g(-2) = (-2)^2

g(-2) = 4

Therefore making

$g(-2) = g(h(5)) = (g \circ h)(5) = 4$

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