Question

Let f(x) = x2 + 4 and g(x) = 3x + 2, evaluate the following.

a. f( - 2) =
b. f( – 1) + g( – 4) =
c. 3f(x) - g(x)=
d. f (g(4))=
e. g(f(4))=

Answer 1

We have two functions:

`\begin{gathered} f(x)=x^2+4 \\ g(x)=3x+2 \end{gathered}`

a) We have to evaluate f(-2). We replace x with -2 and calculate:

`\begin{gathered} f(-2)=(-2)^2+4 \\ f(-2)=4+4 \\ f(-2)=8 \end{gathered}`

b) We have to evaluate f(-1) + g(-4):

`\begin{gathered} f(-1)+g(-4)=\lbrack(-1)^2+4\rbrack+\lbrack3(-4)+2\rbrack \\ f(-1)+g(-4)=\lbrack1+4\rbrack+\lbrack-12+2\rbrack \\ f(-1)+g(-4)=5+(-10) \\ f(-1)+g(-4)=-5 \end{gathered}`

c) In this case,the result will be in function of x:

`\begin{gathered} 3f(x)-g(x)=3(x^2+4)-(3x+2) \\ 3f(x)-g(x)=3x^2+3\cdot4-3x-2 \\ 3f(x)-g(x)=3x^2+12-3x-2 \\ 3f(x)-g(x)=3x^2-3x+10 \end{gathered}`

d) In this case, the argument of f(x) is g(4). We have two ways to solve this:

1. We calculate g(4) and use the result as argument of f(x).

2. We calculate the function f(g(x)) and then use 4 as argument.

We will apply the second method this time:

`\begin{gathered} f(g(x))=\lbrack g(x)\rbrack^2+4 \\ f(g(x))=(3x+2)^2+4 \\ f(g(x))=(3x)^2+2\cdot3x\cdot2+2^2+4 \\ f(g(x))=9x^2+12x+4+4 \\ f(g(x))=9x^2+12x+8 \\ \Rightarrow f(g(4))=9(4)^2+12(4)+8 \\ f(g(4))=9\cdot16+48+8 \\ f(g(4))=144+48+8 \\ f(g(4))=200 \end{gathered}`

e) In this case we have to compose the function, but the order is reversed: f(4) is used as argument of g(x).

We will apply the first method in this case (both methods can be used anytime):

`\begin{gathered} f(4)=4^2+4=16+4=20 \\ \Rightarrow g(f(4))=g(20)=3\cdot20+2=60+2=62 \end{gathered}`

Answer:

a. f(-2) = 8

b. f(-1) + g(-4) = -5

c. 3f(x) - g(x) = 3x² - 3x + 10

d. f(g(4)) = 200

e. g(f(4)) = 62