Function Inverses.pdf - Kuta Software - Free Printable

Let's solve the problem step by step.

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Part 1: State if the given functions are inverses

To determine whether two functions $ f(x) $ and $ g(x) $ are inverses, we need to check if:

$$
f(g(x)) = x \quad \text{and} \quad g(f(x)) = x
$$

If both compositions result in $ x $, then they are inverses.

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#### 1)
$ g(x) = 4 - \frac{3}{2}x $
$ f(x) = \frac{1}{2}x + \frac{3}{2} $

Let’s compute $ f(g(x)) $:

$$
f(g(x)) = f\left(4 - \frac{3}{2}x\right) = \frac{1}{2}\left(4 - \frac{3}{2}x\right) + \frac{3}{2}
= \frac{4}{2} - \frac{3}{4}x + \frac{3}{2} = 2 - \frac{3}{4}x + 1.5 = 3.5 - \frac{3}{4}x
$$

Not equal to $ x $. So not inverses.

But let's double-check carefully:

$$
f(g(x)) = \frac{1}{2}(4 - \frac{3}{2}x) + \frac{3}{2} = 2 - \frac{3}{4}x + \frac{3}{2} = \left(2 + \frac{3}{2}\right) - \frac{3}{4}x = \frac{7}{2} - \frac{3}{4}x \ne x
$$

So no, not inverses.

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#### 2)
$ g(n) = \frac{-12 - 2n}{3} $
$ f(n) = \frac{-5 + 6n}{5} $

We’ll test $ f(g(n)) $:

First, simplify $ g(n) $:
$$
g(n) = \frac{-12 - 2n}{3} = -4 - \frac{2}{3}n
$$

Now plug into $ f $:
$$
f(g(n)) = f\left(-4 - \frac{2}{3}n\right) = \frac{-5 + 6(-4 - \frac{2}{3}n)}{5}
= \frac{-5 -24 -4n}{5} = \frac{-29 - 4n}{5} \ne n
$$

Not an inverse.

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#### 3)
$ f(n) = \frac{-16 + n}{4} $
$ g(n) = 4n + 16 $

Check $ f(g(n)) $:

$$
f(g(n)) = f(4n + 16) = \frac{-16 + (4n + 16)}{4} = \frac{4n}{4} = n
$$

Now $ g(f(n)) $:

$$
g(f(n)) = g\left(\frac{-16 + n}{4}\right) = 4\left(\frac{-16 + n}{4}\right) + 16 = (-16 + n) + 16 = n
$$

✔ Both compositions give $ n $. So yes, inverses.

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#### 4)
$ f(x) = -\frac{4}{7}x - \frac{16}{7} $
$ g(x) = \frac{3}{2}x - \frac{3}{2} $

Try $ f(g(x)) $:

$$
f(g(x)) = f\left(\frac{3}{2}x - \frac{3}{2}\right)
= -\frac{4}{7}\left(\frac{3}{2}x - \frac{3}{2}\right) - \frac{16}{7}
= -\frac{12}{14}x + \frac{12}{14} - \frac{16}{7}
= -\frac{6}{7}x + \frac{6}{7} - \frac{16}{7}
= -\frac{6}{7}x - \frac{10}{7} \ne x
$$

✘ Not inverses.

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#### 5)
$ f(n) = -(n + 1)^3 $
$ g(n) = 3 + n^3 $

Check $ f(g(n)) $:

$$
f(g(n)) = f(3 + n^3) = -(3 + n^3 + 1)^3 = -(n^3 + 4)^3 \ne n
$$

Nope. Try $ g(f(n)) $:

$$
g(f(n)) = g(-(n+1)^3) = 3 + [-(n+1)^3]^3 = 3 - (n+1)^9 \ne n
$$

✘ Not inverses.

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#### 6)
$ f(n) = 2(n - 2)^2 $
$ g(n) = \frac{4 + \sqrt[3]{4n}}{2} $

Wait — cube root? But $ f(n) $ is a square, so it's not one-to-one unless restricted.

But let's try composition.

Note: $ f(n) = 2(n-2)^2 $ → this is a parabola, not one-to-one over all real numbers. So inverse may not exist unless domain is restricted.

But let's try to see if $ f(g(n)) = n $:

$$
g(n) = \frac{4 + \sqrt[3]{4n}}{2}
$$

Then:
$$
f(g(n)) = 2\left(\frac{4 + \sqrt[3]{4n}}{2} - 2\right)^2 = 2\left(\frac{\sqrt[3]{4n}}{2}\right)^2 = 2 \cdot \frac{(\sqrt[3]{4n})^2}{4} = \frac{1}{2} (\sqrt[3]{4n})^2
$$

This is not $ n $. So ✘ not inverses.

Also, note that $ f(n) = 2(n-2)^2 $ has outputs ≥ 0, but $ g(n) $ is defined for all $ n $, so likely not inverses.

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#### 7)
$ f(x) = \frac{4}{-x - 2} + 2 $
$ h(x) = -\frac{1}{x + 3} $

Let’s simplify $ f(x) $:

$$
f(x) = \frac{4}{-x - 2} + 2 = \frac{4}{-(x + 2)} + 2 = -\frac{4}{x+2} + 2
$$

Now compute $ f(h(x)) $:

$$
h(x) = -\frac{1}{x+3}
$$

Plug into $ f $:

$$
f(h(x)) = -\frac{4}{h(x) + 2} + 2 = -\frac{4}{-\frac{1}{x+3} + 2} + 2
= -\frac{4}{\frac{-1 + 2(x+3)}{x+3}} + 2
= -\frac{4}{\frac{-1 + 2x + 6}{x+3}} + 2
= -\frac{4}{\frac{2x + 5}{x+3}} + 2
= -4 \cdot \frac{x+3}{2x+5} + 2
$$

Not equal to $ x $. So ✘ not inverses.

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#### 8)
$ g(x) = -\frac{2}{x} - 1 $
$ f(x) = -\frac{2}{x + 1} $

Try $ f(g(x)) $:

First, $ g(x) = -\frac{2}{x} - 1 = -\left(\frac{2}{x} + 1\right) $

Now plug into $ f $:

$$
f(g(x)) = f\left(-\frac{2}{x} - 1\right) = -\frac{2}{\left(-\frac{2}{x} - 1\right) + 1} = -\frac{2}{-\frac{2}{x}} = -\frac{2}{-2/x} = x
$$

Now check $ g(f(x)) $:

$$
f(x) = -\frac{2}{x+1}
$$

Then:
$$
g(f(x)) = g\left(-\frac{2}{x+1}\right) = -\frac{2}{-\frac{2}{x+1}} - 1 = -\left(-\frac{2(x+1)}{2}\right) - 1 = (x+1) - 1 = x
$$

✔ Both compositions give $ x $. So yes, inverses.

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✔ Summary of Part 1:

| Problem | Are they inverses? |
|--------|--------------------|
| 1) | No |
| 2) | No |
| 3) | Yes |
| 4) | No |
| 5) | No |
| 6) | No |
| 7) | No |
| 8) | Yes |

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Part 2: Find the inverse of each function

To find the inverse, swap $ x $ and $ y $, then solve for $ y $.

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#### 9) $ h(x) = \sqrt[3]{x} - 3 $

Let $ y = \sqrt[3]{x} - 3 $

Swap: $ x = \sqrt[3]{y} - 3 $

Solve for $ y $:

$$
x + 3 = \sqrt[3]{y} \Rightarrow y = (x + 3)^3
$$

So inverse is: $ h^{-1}(x) = (x + 3)^3 $

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#### 10) $ g(x) = \frac{1}{x} - 2 $

Let $ y = \frac{1}{x} - 2 $

Swap: $ x = \frac{1}{y} - 2 $

Solve:

$$
x + 2 = \frac{1}{y} \Rightarrow y = \frac{1}{x + 2}
$$

So $ g^{-1}(x) = \frac{1}{x + 2} $

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#### 11) $ h(x) = 2x^2 + 3 $

This is a quadratic — not one-to-one unless domain is restricted.

But since no restriction is given, we assume general case.

Let $ y = 2x^2 + 3 $

Swap: $ x = 2y^2 + 3 $

Solve for $ y $:

$$
x - 3 = 2y^2 \Rightarrow y^2 = \frac{x - 3}{2} \Rightarrow y = \pm \sqrt{\frac{x - 3}{2}}
$$

So inverse is not a function unless we restrict domain.

But typically, we write:

$$
h^{-1}(x) = \pm \sqrt{\frac{x - 3}{2}}, \quad x \geq 3
$$

But since it's not a function, we say no inverse function exists without domain restriction.

However, if we assume $ x \geq 0 $, then $ h^{-1}(x) = \sqrt{\frac{x - 3}{2}} $

But strictly speaking, no unique inverse unless domain is restricted.

But in most algebra contexts, they expect:

$$
h^{-1}(x) = \pm \sqrt{\frac{x - 3}{2}} \quad \text{(but not a function)}
$$

But better to write:

> The inverse relation is $ h^{-1}(x) = \pm \sqrt{\frac{x - 3}{2}} $, but it's not a function unless the domain of $ h $ is restricted.

But since the question says "find the inverse", and assuming principal branch, perhaps:

$$
\boxed{h^{-1}(x) = \sqrt{\frac{x - 3}{2}}} \quad \text{for } x \geq 3
$$

But technically, only if domain of $ h $ is $ x \geq 0 $.

Alternatively, leave as:

$$
h^{-1}(x) = \pm \sqrt{\frac{x - 3}{2}}, \quad x \geq 3
$$

But for simplicity, often written as:

$$
\boxed{h^{-1}(x) = \sqrt{\frac{x - 3}{2}}}
$$

with understanding of domain.

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#### 12) $ g(x) = -4x + 1 $

Linear function — easy.

Let $ y = -4x + 1 $

Swap: $ x = -4y + 1 $

Solve:

$$
x - 1 = -4y \Rightarrow y = \frac{1 - x}{4}
$$

So $ g^{-1}(x) = \frac{1 - x}{4} $

Or $ g^{-1}(x) = -\frac{1}{4}x + \frac{1}{4} $

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✔ Final Answers:

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Part 1: Are they inverses?

1) No
2) No
3) Yes
4) No
5) No
6) No
7) No
8) Yes

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Part 2: Inverses

9) $ h^{-1}(x) = (x + 3)^3 $

10) $ g^{-1}(x) = \frac{1}{x + 2} $

11) $ h^{-1}(x) = \pm \sqrt{\frac{x - 3}{2}} $, or more precisely, $ h^{-1}(x) = \sqrt{\frac{x - 3}{2}} $ if domain of $ h $ is $ x \geq 0 $

12) $ g^{-1}(x) = \frac{1 - x}{4} $

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Let me know if you want graphs or further explanation!