We will begin by finding the inverse of

f (x) .

First, we need to replace

f (x)

with

y .

From there, we switch

x

and

y

and solve for

y .

y = \frac{1}{- 2 x} \to x = \frac{1}{- 2 y}

The resulting equation will be the inverse of the given function.

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

▼

Solve for

y

RaiseEqn

$LHS^{2} = RHS^{2}$

x^{2} = \frac{1}{- 2 y}

MultEqn

$LHS \cdot y = RHS \cdot y$

y x^{2} = \frac{1}{- 2}

DivEqn

$LHS / x^{2} = RHS / x^{2}$

y = \frac{1}{- 2 x ^{2}}

MoveNegDenomToFrac

Put minus sign in front of fraction

y = - \frac{1}{2 x ^{2}}

Finally, to indicate that this is the inverse function of

f (x),

we will replace

y

with

f^{- 1} (x) .

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

Now that we found the inverse of

f (x),

we will determine the domain and range of

f (x)

and

f^{- 1} .

Domain and Range of the Function

To determine the domain of the given function, the radicand cannot be negative and the denominator of the quotient cannot be

0 .

This means that the radicand must be greater than

0 .

Let's find the value(s) of

x

that make this true.

- 2 x > 0

DivNegIneq

Divide by $- 2$ and flip inequality sign

x < 0

All values of

x

that are less than

0

are included in the domain.

Domain : x < 0

For the range, think about the fact that the principal root of a number is always positive.

f (x) = \frac{1}{- 2 x} \leftarrow \leftarrow positive positive

Because the quotient of two positive numbers is always a positive number, the range is all values of

y

such that

y > 0 .

Range : y > 0

Domain and Range of the Inverse

When we find the inverse of a function, we are basically exchanging its

x

and

y

values. Therefore, the range of

f (x)

becomes the domain of

f^{- 1} (x),

and the domain of

f (x)

becomes the range of

f^{- 1} (x) .

Domain : x > 0 Range : y < 0

Is the Inverse a Function?

A function is a relation where each input is related to exactly one output. In this case, for each $x$ in domain of $f^{- 1} (x),$ there is only one value of $y$ in the range. Therefore, $f^{- 1} (x)$ is a function in its domain.

Exercise

This Solution is Missing. Mathvizy Can Solve This Problem For You