Question #c8396

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Answer 1 · 2017-12-30T11:10:59

#h(g(x))=5x^2-8#
#h(g(-3))=37#

I'll start with #(hcircg)(x)=h(g(x))#:

#h(x)=5x-3# and #g(x)=x^2-1# so replace every #x# in #h(x)# with #x^2-1#. #h(g(x))=5(x^2-1)-3=5x^2-5-3=5x^2-8#.

So #h(g(x))=5x^2-8#.

#h(g(-3))# can be found two ways. First find #g(-3)#:

#g(-3)=(-3)^2-1=9-1=8#

So we're finding #h(8) = 5(8)-3 = 40-3=37#.

Alternatively, we could use our simplified #h(g(x))#:

#h(g(-3))=5(-3)^2-8=5*9-8=45-8=37#.