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Radical functions can be combined, through addition, subtraction and multiplication. Depending on the original functions and which operations are used, the resulting function can have similar or new characteristics.
Exercise

Consider the functions. $f(x)=\sqrt{x} \quad \text{ and } \quad g(x)=\sqrt{4x}$ Find $h(x)$ if $h(x)=f(x) \cdot g(x).$

Solution
When functions containing one or more radicals are multiplied, it is often helpful to rewrite the expressions using rational exponents.
$h(x)=f(x) \cdot g(x)$
$h(x)={\color{#0000FF}{\sqrt{x}}} \cdot {\color{#009600}{\sqrt{4x}}}$
$h(x)={x}^{\frac{1}{4}} \cdot \sqrt{4x}$
$h(x)={x}^{\frac{1}{4}} \cdot (4x)^{\frac{1}{2}}$
Next, the properties of exponents can be used to simplify the product.
$h(x)={x}^{\frac{1}{4}} \cdot (4x)^{\frac{1}{2}}$
$h(x)={x}^{\frac{1}{4}} \cdot 4^{\frac{1}{2}} \cdot x^{\frac{1}{2}}$
$h(x)={x}^{\frac{1}{4}} \cdot 2 \cdot x^{\frac{1}{2}}$
$h(x)=2 {x}^{\frac{1}{4}} \cdot x^{\frac{1}{2}}$
$h(x)=2 {x}^{\frac{1}{4} + \frac{1}{2}}$
$h(x)=2 {x}^{\frac{1}{4} + \frac{2}{4}}$
$h(x)=2 {x}^{\frac{3}{4}}$
Now that the function $h(x)$ has been simplified completely, we can rewrite it with radical. $h(x)=2 {x}^{\frac{3}{4}} \quad \Leftrightarrow \quad h(x)=2 \sqrt{x^3}$
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Concept

## Inverse Functions

Inverse functions are two functions that undo each other. The functions $f(x)$ and $g(x)$ are inverses of each other if $f\left(g(x)\right)=x \quad \text{and} \quad g\left(f(x)\right)=x.$

When a function is denoted $f(x),$ its inverse is often referred to as $f^{\text{-} 1}(x).$
Concept

## Inverse Functions to Radical Functions

Radicals can be eliminated by raising them to the same power as the index of the radical. Therefore, the inverse function to a radical function is typically a power function. The radical function $f(x)=\sqrt[n]{x}$ has the inverse function $f^{\text{-} 1}(x)=x^n.$ This can be proven by showing $\textbf{I:}\ f(f^{\text{-} 1}(x))=x \quad \text{and} \quad \textbf{II:}\ f^{\text{-} 1}(f(x))=x.$ Consider I.

### Rule

info
$f(f^{\text{-} 1}(x))=x$
$f(x)=\sqrt[n]{x}$
$f({\color{#0000FF}{f^{\text{-} 1}(x)}}) = \sqrt[n]{{\color{#0000FF}{f^{\text{-} 1}(x)}}}$
$f(f^{\text{-} 1}(x))=\sqrt[n]{{\color{#0000FF}{x^n}}}$
$f(f^{\text{-} 1}(x))=x$
Next, consider II.

### Rule

info
$f^{\text{-} 1}(f(x))=x$
$f^{\text{-} 1}(x)=x^n$
$f^{\text{-} 1}({\color{#0000FF}{f(x)}})=({\color{#0000FF}{f(x)}})^n$
$f^{\text{-} 1}(f(x))=(\sqrt[n]{x})^n$
$\left ( \sqrt[n]{a}\right )^n = a$
$f^{\text{-} 1}(f(x))=x$
It has been shown that $\sqrt[n]{x}$ and $x^n$ are inverses of each other. This is true for all values of $n.$
Exercise

Determine if the two functions are inverses of each other. $f(x)=1+\sqrt{x} \quad \text{and} \quad g(x)=(x-1)^3$

Solution

The two functions, $f(x)$ and $g(x),$ are inverses if $f(g(x))=x \quad \text{and} \quad g(f(x))=x.$ Thus, if we can show that

1. $f(g(x))=x$, and
2. $g(f(x))=x$
we can conclude that $f$ and $g$ are inverses. We'll begin with $f(g(x)).$
$f(x)=1+\sqrt{x}$
$f\left({\color{#0000FF}{g(x)}} \right)=1+\sqrt{{\color{#0000FF}{g(x)}}}$
$f(g(x))=1+\sqrt{{\color{#0000FF}{(x-1)^3}}}$
$f(g(x))=1+(x-1)$
$f(g(x))=1+x-1$
$f(g(x))=x$
Now that we've satisfied $1,$ we'll attempt to satisfy $2.$
$g(x)=(x-1)^3$
$g({\color{#0000FF}{f(x)}})=\left( {\color{#0000FF}{f(x)}}-1 \right)^3$
$g(f(x))=\left( \left({\color{#0000FF}{1+\sqrt{x}}}\right)-1 \right)^3$
$g(f(x))=\left( 1+\sqrt{x}-1 \right)^3$
$g(f(x))=\left( \sqrt{x} \right)^3$
$\left ( \sqrt[n]{a}\right )^n = a$
$g(f(x))=x$
Since both $f(g(x))=x \quad \text{and} \quad g(f(x))=x,$ $f$ and $g$ are inverses of each other.
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Exercise

The graph of the function $f(x)=2\sqrt{x}$ is shown. Use the graph to show that $f(x)$ is invertible. Then, determine the inverse, $f^{\text{-} 1}(x),$ and graph both functions.

Solution

Graphically, it can be shown that a function is invertible by performing the horizontal line test. If we move an imaginary horizontal line across the graph of a function, and the line intersects the graph more than once anywhere, the function is not invertible. Here, we'll draw more than one line to be sure. The horizontal lines in the graph both intersect the function once. If we place a line anywhere else it still intersects the graph at the most once. Therefore, the function is invertible. Next, we can find the inverse, $f^{\text{-} 1}(x)$ algebraically. To begin, replace $f(x)$ with $y$ in the given rule. $y=2\sqrt{x}$ Next step is to switch $x$ and $y$ in the function rule. $y=2\sqrt{x} \quad \stackrel{\text{switch}}{\longrightarrow} \quad x=2\sqrt{y}$ Now we need to solve for $y.$ The resulting equation will be the inverse of the given function.
$x=2\sqrt{y}$
$\dfrac{x}{2}=\sqrt{y}$
$\left( \dfrac{x}{2} \right)^6=\left( \sqrt{y} \right)^6$
$\left ( \sqrt[n]{a}\right )^n = a$
$\left( \dfrac{x}{2} \right)^6=y$
$y=\left( \dfrac{x}{2} \right)^6$
Thus, the inverse is $f^{\text{-} 1}(x)=\left( \dfrac{x}{2} \right)^6.$ We'll graph $f$ and $f^{\text{-} 1}$ in the same coordinate plane. The domain of $f(x)$ is restricted to $x\geq 0.$ As a consequence, the domain of $f^{\text{-} 1}$ must be restricted to the same interval, even though $\left(\frac{x}{2}\right)^6$ is defined for all values of $x.$

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