How do you find #(g" o" f)(2)# given #f(x)=-2x+1# and #g(x)=sqrt(x^2-5)#?

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Answer 1 · 2016-08-19T04:35:36

#(g@f)(2) = 2#

By function compositions:

#g(f(x)) = sqrt((-2x + 1)^2 - 5)#

#g(f(x)) = sqrt(4x^2 - 4x + 1 - 5)#

#g(f(x)) = sqrt(4x^2 - 4x - 4)#

#g(f(x)) = sqrt(4(x^2 - x - 1))#

#g(f(x)) = 2sqrt(x^2 - x - 1)#

#g(f(2)) = 2sqrt(2^2 - 2 - 1)#

#g(f(2)) = 2(1)#

#g(f(2)) = 2#

Hopefully this helps!