Steply

Inverse of f(x) = √(x + 3)

Step by step, with the reason for every step · Inverses of non-linear functions · Algebra 2

Step by step

The question

Question: find the inverse of f(x) = √(x + 3).

f(x) = √x + 3

The inverse goes back: what comes out of f goes in, and what went in comes out.

inverse: swaps in and out

1 Write y =

First we write y for f(x): y = √(x + 3).

y = √x + 3

2 Swap x and y

Swap x and y: what comes out now goes in.

x = √(y + 3).

x = √y + 3

3 Get y alone

To remove the square root, we square both sides.

Squaring removes the square root.

(√9)² = 9: squaring removes the √

Square both sides: (x)² = y + 3.

(x)2 = y + 3

To remove + 3, we take away 3.

We do it on both sides, so they stay equal.

x2 − 3 = y + 3 − 3

+ 3 and − 3 make zero.

Cross them out.

y is alone: y = x² − 3.

y = x2 − 3

This y is the inverse: f⁻¹(x) = x² − 3.

f −1(x) = x2 − 3, x ≥ 0

4 The domain of the inverse

The inverse's x's are f's answers.

f(x) is a square root , so it is 0 or more.

f's y's = the inverse's x's

Careful!

Without it, the inverse would give answers f never had.

Its x's are only f's answers: 0 or more.

Careful: keep x ≥ 0 for the inverse

So the inverse only takes x ≥ 0.

x ≥ 0

5 Draw both

Draw the grid.

Plot points of f: (−3, 0), (−2, 1) and (1, 2).

Join them: f(x) = √(x + 3), starting at (−3, 0).

For the inverse, swap each point: (0, −3), (1, −2) and (2, 1).

Join them: f⁻¹, starting at (0, −3).

The two graphs are mirror images in the line y = x.

6 Check

Let's check: there and back.

f(1) = 2.

f(1) = √1 + 3 = 2

Now put 2 into the inverse.

We get back 1.

It works!

f −1(2) = (2)2 − 3 = (2)2 − 3 = 1

Let's look back.

First: write y =.

Then: swap x and y.

Then: get y alone.

Then: the domain of the inverse.

Then: draw both.

Last: check.

Answer
f −1(x) = x2 − 3, x ≥ 0

Your turn

Try this one yourself, then check:

inverse of f(x) = √x + 1
Show the answer
f −1(x) = x2 − 1, x ≥ 0

More from Inverses of non-linear functions

The whole lesson: Inverses of non-linear functions →