6.3: Radical Functions

From LibreTexts, under CC BY-NC-SA 4.0, as the source states it; its equations are written in TeX, which the PDF shows typeset. This page at LibreTexts.

Radical functions are the most famous example of an inverse function.

For $f(x)=x^n$, we say that $f^{-1}(x)=x^{\frac{1}{n}}=\sqrt[n]{x}$

When $n$ is an odd positive integer, $f(x)$ is 1-to-1 and the domain of $f^{-1}$ is $(-\infty,\infty)$:

$clipboard_ea71e2a11331287c9b41d59bdfa0de17d.png$

But when $n$ is an even positive integer, $f(x)$ fails the horizontal line test. To create an inverse for these functions, we restrict the domain of $f(x)$ to $[0,\infty)$, this results in $f^{-1}(x)$ existing and having domain $[0,\infty)$ as well:

$clipboard_ea3ea0117eae697d9a7b53964923aaba6.png$

While we have already seen and used radicals previously (such as in the quadratic formula), we have not yet investigated them as functions.

From the graphs we can see that they are clearly continuous on their domain.

This generalizes to $\displaystyle{\lim_{x\to a} \sqrt[n]{g(x)}=\sqrt[n]{\lim_{x\to a}g(x)}}$ (whenever the radical makes sense)

Remember: radicals are relatively well behaved when it comes to multiplication and division:

$\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}$ and $\displaystyle{\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}}$ whenever all terms exist.

But they almost never distribute across a plus or minus sign:

$\sqrt[n]{a+b}\ne \sqrt[n]{a}+\sqrt[n]{b}$ and $\sqrt[n]{a-b}\ne \sqrt[n]{a}-\sqrt[n]{b}$

Sometimes radicals end up in places that make us uncomfortable (like as part of a fraction) and we want to re-write them in a different way. This can be done by using a difference of $n^{th}$'s formula:

$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +a^1b^{n-1}+b^{n-1}\right)$

when $n=2$, this is called the difference of squares $a^2-b^2=(a-b)(a+b)$

These can be used to rationalize (get rid of radicals in) a denominator or numerator if desired:

Example $\PageIndex{1}$

(a) Rationalize the denominator for $f(x)=\displaystyle{ \frac{x+1}{\sqrt{x+2}-3}}$

(b) Rationalize the numerator for $g(x)=\displaystyle{ \frac{\sqrt[3]{x}-2}{x-8}}$

Solution

(a) To eliminate $\sqrt{x+2}$, we multiply top and bottom by $\sqrt{x+2}+3$

$f(x)=\displaystyle{ \frac{x+1}{\sqrt{x+2}-3}\cdot \textcolor{red}{\frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}}=\frac{(x+1)(\sqrt{x+2}+3)}{(\sqrt{x+2})^2-3^2}=\frac{(x+1)(\sqrt{x+2}+3)}{(x+2)-9}= \frac{(x+1)(\sqrt{x+2}+3)}{x-7}}$

(b) To eliminate $\sqrt[3]{x}$, we multiply top and bottom by $(\sqrt[3]{x})^2+(\sqrt[3]{x})\cdot 2+2^2$

$\displaystyle{ \frac{(\sqrt[3]{x})^3-2^3}{(x-8)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}}=\displaystyle{ \frac{x-8}{(x-8)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}=\frac{1}{ \sqrt[3]{x^2}+2\sqrt[3]{x}+4}}$ (for $x\ne 8$)

As we can see above, sometimes rationalizing a numerator or denominator can help with reducing, which means it is definitely a great tool for taking limits.

Evaluate the following limits involving radical functions:

Example $\PageIndex{2}$

Evaluate each of the limits below by rationalizing the numerator or denominator.

(a) $ \displaystyle{ \lim_{x\to 0} \frac{x}{\sqrt{x+1}-1}}$

(b) $ \displaystyle{ \lim_{t\to 4} \frac{\sqrt{t^2+9}-5}{t-4}}$

Solution

(a) $\displaystyle{ \lim_{x\to 0} \frac{x}{\sqrt{x+1}-1}=\lim_{x\to 0} \frac{x}{\sqrt{x+1}-1}\cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\lim_{x\to 0} \frac{x(\sqrt{x+1}+1)}{(\sqrt{x+1})^2-1^2}}$

$\displaystyle{ =\lim_{x\to 0} \frac{x(\sqrt{x+1}+1)}{x} =\lim_{x\to 0} \frac{\sqrt{x+1}+1}{1} = \frac{\sqrt{0+1}+1}{1}= 2}$

(b) $\displaystyle{ \lim_{t\to 4} \frac{\sqrt{t^2+9}-5}{t-4}=\lim_{t\to 4} \frac{\sqrt{t^2+9}-5}{t-4}\cdot \frac{\sqrt{t^2+9}+5}{\sqrt{t^2+9}+5}=\lim_{t\to 4} \frac{(\sqrt{t^2+9})^2-5^2}{(t-4)(\sqrt{t^2+9}+5)}}$

$ \displaystyle{ =\lim_{t\to 4} \frac{t^2-16}{(t-4)(\sqrt{t^2+9}+5)} =\lim_{t\to 4} \frac{(t-4)(t+4)}{(t-4)(\sqrt{t^2+9}+5)} =\lim_{t\to 4} \frac{t+4}{\sqrt{t^2+9}+5}= \frac{4+4}{\sqrt{4^2+9}+5}=\frac{4}{5}} $

(c) $\displaystyle{ \lim_{y\to -2} \frac{y^2+9y+8}{\sqrt[3]{y}+2}=\lim_{y\to -8} \frac{(y+8)(y+1)}{\sqrt[3]{y}-(-2)}\cdot \frac{\sqrt[3]{y^2}+(-2)\sqrt[3]{y}+(-2)^2}{\sqrt[3]{y^2}+(-2)\sqrt[3]{y}+(-2)^2 }}$

$ \displaystyle{ =\lim_{y\to -8} \frac{(y+8)(y+1)\left(\sqrt[3]{y^2}-2\sqrt[3]{y}+4\right)}{\left(\sqrt[3]{y}\right)^3-(-2)^3} =\lim_{y\to -8} \frac{(y+8)(y+1)\left(\sqrt[3]{y^2}-2\sqrt[3]{y}+4\right)}{y+8} }$

$\displaystyle{ =\lim_{y\to -8} (y+1)\left(\sqrt[3]{y^2}-2\sqrt[3]{y}+4\right) = (-8+1)\left(\sqrt[3]{(-8)^2}-2\sqrt[3]{-8}+4\right) = -7\left(4+4+4\right)=-84 }$

Remember that $\sqrt[n]{x^n}=x$ when $n$ is odd, but $\sqrt[n]{x^n}=|x|$ when $n$ is even!

This can make a big difference when using the Lebron James Rule:

Example $\PageIndex{3}$

(a) $\displaystyle{ \lim_{x\to \infty } \frac{\sqrt{9x^2+4}+5x}{2x+7}}$

(b) $\displaystyle{ \lim_{x\to -\infty } \frac{\sqrt{9x^2+4}+5x}{2x+7}}$

Solution

(a) Get rid of terms that are clearly smaller

$\displaystyle{\lim_{x\to \infty } \frac{\sqrt{9x^2+4}+5x}{2x+7}= \lim_{x\to \infty } \frac{\sqrt{9x^2}+5x}{2x}= \lim_{x\to \infty } \frac{|3x|+5x}{2x}}$

Since $x\to +\infty$ we know $x>0$ so $|x|=x$, turning our limit into:

$\displaystyle{= \lim_{x\to \infty } \frac{3x+5x}{2x}= \lim_{x\to \infty } \frac{8x}{2x}= \lim_{x\to \infty } 4=4}$

(b) $\displaystyle{ \lim_{x\to -\infty } \frac{\sqrt{9x^2+4}+5x}{2x+7}=\lim_{x\to -\infty } \frac{\sqrt{9x^2}+5x}{2x} =\lim_{x\to -\infty } \frac{|3x|+5x}{2x}}$

Since $x\to -\infty$ we know $x<0$ so $|x|=-x$, turning our limit into:

$\displaystyle{ =\lim_{x\to -\infty } \frac{-3x+5x}{2x}=\lim_{x\to -\infty } \frac{2x}{2x}=\lim_{x\to -\infty } 1=1}$

Just remember, whenever the limit is $\infty - \infty$ and youre not sure who is biggest, you should rewrite it first:

Example $\PageIndex{4}$

$\displaystyle{ \lim_{w\to \infty } \sqrt{4w^2+2w-1}-2w}$

Solution

Tempting to write as $\sqrt{4w^2}-2w$, but this would be wrong!
$\displaystyle{ \lim_{w\to \infty } \sqrt{4w^2+2w-1}-2w=\lim_{w\to \infty } \left(\sqrt{4w^2+2w-1}-2w\right) \cdot \frac{\sqrt{4w^2+2w-1}+2w}{\sqrt{4w^2+2w-1}+2w}}$

$\displaystyle{ =\lim_{w\to \infty } \frac{\left(\sqrt{4w^2+2w-1}\right)^2-(2w)^2}{\sqrt{4w^2+2w-1}+2w} =\lim_{w\to \infty } \frac{2w-1}{\sqrt{4w^2+2w-1}+2w}}$

Now use the Lebron James rule

$\displaystyle{ =\lim_{w\to \infty } \frac{2w}{\sqrt{4w^2}+2w}=\lim_{w\to \infty } \frac{2w}{|2w|+2w}}\displaystyle{ =\lim_{w\to \infty } \frac{2w}{2w+2w}=\lim_{w\to \infty } \frac{1}{2}=\frac{1}{2}}$

This page titled 6.3: Radical Functions was last modified on Thu, 13 Mar 2025 20:04:46 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kenn Huber.

Contents

Text

Example \(\PageIndex{1}\)

Solution

Example \(\PageIndex{2}\)

Solution

Example \(\PageIndex{3}\)

Solution

Example \(\PageIndex{4}\)

Solution