# Help check my calculation? "If f(x) = 2x - 3 & g(x) = x^2+5, find (a) gf(2) = g(f(2)), (b) fg(-3) = f(g(-3))

#### Jacob

##### New member
If f(x) = 2x - 3 & g(x) = x^2+5, find:

a) gf(2) = g(f(2))
= g(2(2)-3)
= 2x - 3 (4 - 3)
= 2x - 12 + 9
= 2x - 3 #

b) fg(-3) = f(g(-3))
= f(-3^2+5)
= 2x - 3 ( -9 + 5)
= 2x + 27 - 15
= 2x + 12 #

Am I doing it correctly?

#### Subhotosh Khan

##### Super Moderator
Staff member
If f(x) = 2x - 3 & g(x) = x^2+5, find:

a) gf(2) = g(f(2))
= g(2(2)-3)
= 2x - 3 (4 - 3)
= 2x - 12 + 9
= 2x - 3 #..........................Incorrect - see below

b) fg(-3) = f(g(-3))
= f(-3^2+5)
= 2x - 3 ( -9 + 5)
= 2x + 27 - 15
= 2x + 12 #..........................Incorrect - see below

Am I doing it correctly?
If f(x) = 2x - 3 & g(x) = x^2+5, find:

a) gf(2) = g(f(2))

f(2) = 2*2 - 3 = 1

gf(2) = g(f(2)) = g(1) = 1^2 + 5 = 6

b) fg(-3) = f(g(-3))

g(-3) = (-3)^2 + 5 = 9 +5 = 14

fg(-3) = f(g(-3)) = f(14) = 2*14 - 3 = 25

#### JeffM

##### Elite Member
If f(x) = 2x - 3 & g(x) = x^2+5, find:

a) gf(2) = g(f(2))
= g(2(2)-3)
= 2x - 3 (4 - 3)
= 2x - 12 + 9
= 2x - 3 #

b) fg(-3) = f(g(-3))
= f(-3^2+5)
= 2x - 3 ( -9 + 5)
= 2x + 27 - 15
= 2x + 12 #

Am I doing it correctly?
No. A composition of functions is itself a function. You feed the composition a number, and it spits out a number.

One obvious way to do this kind of problem is to apply the function rules from the inside out as required by PEMDAS.

$$\displaystyle f(2) = 2(2) - 3 = 1.$$

$$\displaystyle \therefore g(f(2)) = g(1) = 1^2 + 5 = 6.$$

Now you try the second problem and tell us what you get?

#### Jomo

##### Elite Member
If f(x) = 2x - 3 & g(x) = x^2+5, find:

a) gf(2) = g(f(2))
= g(2(2)-3)
= 2x - 3 (4 - 3)
= 2x - 12 + 9
= 2x - 3 #

b) fg(-3) = f(g(-3))
= f(-3^2+5)
= 2x - 3 ( -9 + 5)
= 2x + 27 - 15
= 2x + 12 #

Am I doing it correctly?
A quick way to know you are wrong is because g(f(2)) must be a pure number, not an expression involving x.

#### HallsofIvy

##### Elite Member
As Jomo said, "gf(2)" has no "x" so its value cannot depend on x! $$\displaystyle f(x)= 2x- 3$$ so f(2)= 2(2)- 3= 4- 3= 1. $$\displaystyle g(x)= x^2+ 5$$ so $$\displaystyle g(f(2))= g(1)= 1^2+ 5= 1+ 5= 6$$.