college algebra

Question

Given f(x) = x2 + 1 and g(x) = x + 3. Calculate the following:

a) (f + g)(2) b) (f − g)(2) c) (f · g)(2) d) (f)(2) g e) f(g(2)) f) g(f(2))

Answer 1

Oh Jennifer:

 

 (a)  for the sum Let S(x) = (F(x)+g(x)) = x^2 + 1 + x+3 = x^2 + x + 4,

        so S(2)= (f+g)(2) = 2^2 + 2 + 4 = 4 + 2 + 4= 10

 

 (b) for the difference Let D(x) = x^2+1 - (x+3) = x^2 + 1 - x - 3= x^2 - x -2,

        so  D(2) = (f-g)(2) = 2^2 - 2 - 2 = 4 - 2- 2 = 0

 

(c) for the product, Let M(x) = (f * g)(x) = (x^2+1)(x+3) = x^3 + 3X^2 + x + 3,

      so M(2) = (f*g)(2) = 2^3 + 3*2^2 + 2+ 3 = 8 + 3*4 + 2 +3 <-- order of operatios PEMDAS

                                                                    = 8 + 12 +2 +3 = 25

 

Note that for parts (a)-(c), you could have found f(2) = 2^2+1 = 5 while g(2) = 2+3 = 5 and then

performed the operations on them so that for (a) (f+g)(2) = f(2)+g(2) = 5 + 5 = 10

                                                                  (b) (f-g)(2) = f(2)-g(2) = 5- 5 = 0

                                                                  (c) (f*g)(2) = f(2)*g(2) = 5*5 = 25

However, this shortcut does not work with parts (d)-(f) because they are function composition!

Please clarify part (d). 

 

Part (e):   f(g(2)) = (x+3)^2  + 1, so f(g(2)) = (2+3)^2 + 1 = 5^2 + 1 = 25 + 1 = 26.

                Likewise, since g(2) = 5, then f(g(2)) = f(5) = 5^2+1 = 25+1 = 26

 

Part (f): g(f(x)) = (x^2+1)+3 = x^2 +4, so g(f(2)) = 2^2+4 = 4+4 = 8. 

            Likewise, since f(2)=5, then g(f(2)) = g(5) = 5 +3 = 8.