Let #f(x)=2x+3# and #g(x)=x^2-4# and #h(x)=x-3/2#, how do you find g(f(3))?

2 Answers
May 11, 2015

The answer is : #g(f(3)) = 77#

Firstly, let's find #g(f(x))# :

You have #f(x) = 2x+3# and #g(x) = x^(2) - 4#, so

#g(f(x)) = f^(2)(x) -4 = (2x+3)^(2) -4#

#g(f(x)) = 4x^(2)+12x+9-4=4x^(2)+12x+5#

Now you can calculate #g(f(3))# :

#g(f(3)) = 4*3^(2) + 12*3 + 5 = 36 + 36 + 5 = 77#

May 11, 2015

Have a look. I think it is ok, although I didn't used #h(x)#...
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