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

Answer 1

To find (g" o" f)(2), where f(x) = -2x + 1 and g(x) = sqrt(x^2 - 5), we first find f(2), then substitute the result into g(x), and finally evaluate g(f(2)).

Find f(2): f(2) = -2(2) + 1 = -4 + 1 = -3.
Substitute f(2) into g(x): g(-3) = sqrt((-3)^2 - 5) = sqrt(9 - 5) = sqrt(4) = 2.

Therefore, (g" o" f)(2) = 2.

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Answer 2

Julia Adams

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

Through compositions of functions:

#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#

I hope this is useful!

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.