3 years ago

Let g(x)=x-3 and h(x)=x^2+6 find (h o g) (1)

2 answers:

5 0

$g(x)=x-3 \ \ \ and\ \ \ h(x)=x^2+6 \\\\(h \circ g) (1)=h\bigg (g(1)\bigg)=h\bigg(1-3\bigg)=h(-2)=(-2)^2+6=4+6=10$

Rudiy273 years ago

5 0

$h \circ g=(x-3)^2+6=x^2-6x+9+6=x^2-6x+15$

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