Steply

f(x) = x - 1, g(x) = 2x, find f(g(x)) and g(f(x))

Step by step, with the reason for every step · Composition order · Algebra 2

Step by step

The question

Question: find f(g(x)) and g(f(x)) for f(x) = x − 1 and g(x) = 2x.

f(x) = x − 1
g(x) = 2x

1 Find f(g(x))

f(g(x)) is a function inside a function.

Like with brackets, we start inside.

First g(x).

inside first: g(x)

g(x) is 2x.

So f(g(x)) is f(2x).

f(g(x)) = f(2x)

Put (2x) in place of every x in f.

We keep it in brackets.

= (2x) − 1

A bracket with a plus or nothing in front can just be removed.

( ) away

Now we have 2x − 1.

f(g(x)) = 2x − 1

2 Find g(f(x))

g(f(x)) is a function inside a function.

Like with brackets, we start inside.

First f(x).

f(x) is x − 1.

So g(f(x)) is g(x − 1).

g(f(x)) = g(x − 1)

Put (x − 1) in place of every x in g.

We keep it in brackets.

= 2(x − 1)

The 2 in front of the bracket multiplies every part inside.

a(b + c) = ab + ac

So we write 2 times each part.

= 2(x) + 2(−1)
− × + = −

Watch the signs.

2 times x is 2x.

2 times (−1) is −2.

g(f(x)) = 2x − 2

3 Compare

2x − 1 and 2x − 2 are not the same.

So the order matters!

2x − 1 ≠ 2x − 2

Careful!

Inside first, then outside.

Swap them, and you usually get a different rule.

Careful: f(g(x)) is usually not g(f(x))

4 Check

Let's check with a number: x = 1.

g of 1 is 2, then f of 2 is 1.

f(g(1)) = f(2) = 1

f of 1 is 0, then g of 0 is 0.

Different, as we found.

g(f(1)) = g(0) = 0

And our rules give the same numbers: 2x − 1 gives 1, 2x − 2 gives 0.

They are right!

2x − 1 → 1, 2x − 2 → 0

Let's look back.

First: find f(g(x)).

Then: find g(f(x)).

Then: compare.

Last: check.

Answer
2x − 1 ≠ 2x − 2

Your turn

Try this one yourself, then check:

f(x) = 3x − 2, g(x) = x, find f(g(x)) and g(f(x))
Show the answer
3x − 2 = 3x − 2

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