Tag: Inverses

Find an inverse for a function defined by an integral

by RoRi

December 20, 2015

Define a function for $x \geq 0$ by

$f(x) = \int_0^x (1+t^3)^{-\frac{1}{2}} \, dt.$

Prove that $f(x)$ is strictly increasing on the nonnegative real axis.
If $g$ denotes the inverse of $f$ prove that $g''$ is proportional to $g^2$ and find the constant of proportionality.

Proof. To show $f(x)$ is strictly increasing we take the derivative,
$f(x) = \int_0^x \frac{1}{\sqrt{1+t^3}} \, dt \quad \implies \quad f'(x) = \frac{1}{\sqrt{1+x^3}}.$

Since $f'(x) > 0$ for all $x \geq 0$ we have that $f(x)$ is strictly increasing on the nonnegative real axis $. \qquad \blacksquare$
Proof. If $g$ is the inverse of $f$ then we know (Theorem 6.7 on page 252 of Apostol)
$g'(y) = \frac{1}{f'(x)}.$

But, we have defined $g$ to the function such that $g(f(x)) = x$ . Therefore,

$g'(y) = \frac{1}{f'(g(y))}.$

Therefore, we have that

$f'(x) = \frac{1}{\sqrt{1+x^3}} \quad \implies \quad g'(y) = \sqrt{1+g^3(y)}.$

Taking another derivative of $g$ with respect to $y$ we then have

$\begin{align*} g''(y) &= \frac{3 (g(y))^2 \cdot g'(y)}{2 \sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} \cdot (g(y))^2 \cdot \frac{g'(y)}{\sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} \cdot (g(y))^2 \cdot \frac{\sqrt{1+(g(y))^3}}{\sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} (g(y))^2. \qquad \blacksquare \end{align*}$

Find an inverse for the function log |x|

by RoRi

December 20, 2015

Consider the function $f(x) = \log |x|$ for $x < 0$ . Prove that this function has an inverse, determine the domain of this inverse, and find a formula to compute the inverse $g(y)$ .

Proof. From the discussion on page 146 of Apostol we know that a function which is continuous and strictly monotonic on an interval $[a,b]$ has an inverse on $[a,b]$ . The function $f(x) = \log |x|$ is continuous and strictly monotonic on the negative real axis; therefore, it has an inverse. We know it is continuous since the log function is continuous on the positive real axis, and $|x| > 0$ for all $x$ , in particular, for all $x < 0$ . Furthermore, we know it is strictly monotonic since

$f'(x) = \frac{1}{x} < 0 \qquad \text{for all } x < 0.$

Therefore, $f(x) = \log |x|$ has an inverse for all $x < 0$ . The domain of this inverse is the range of $\log |x|$ which is all of $\mathbb{R}. \qquad \blacksquare$

To find a formula for the inverse we set

$y = \log |x| \quad \implies \quad e^y = |x| \quad \implies \quad x = -e^y.$

Therefore, $g(y) = -e^y$ valid for all $y \in \mathbb{R}$ .

A sketch for the graph of $g$ is given by

Rendered by QuickLaTeX.com

Establish the formula for the derivative of arccsc x

by RoRi

November 29, 2015

Establish the following formula for the derivative of $\arcsec x$ is correct,

$D(\operatorname{arccsc} x) = \frac{-1}{|x|\sqrt{x^2-1}}, \qquad \text{for all } |x| >1.$

For $x > 1$ let

$y = \operatorname{arccsc} x \quad \implies \quad x = \csc y.$

Then we know

$\frac{dx}{dy} = -\csc y \cot y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\csc y \cot y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\csc y \cot y} \\ && \implies && D(\operatorname{arccsc} x) &= \frac{-1}{\csc y \cot y} \\ && \implies && (D(\operatorname{arccsc} x))^2 &= \left( \frac{-1}{\csc y \cot y} \right)^2 \\ && \implies && (D(\operatorname{arccsc} x))^2 &= \frac{1}{\csc^2 y \cot^2 y} \\ && \implies && (D(\operatorname{arccsc} x))^2 &= \frac{1}{\csc^2 y (\csc^2 y - 1)}. \end{align*}$

Where we use the trig identity $\cot^2 y = \csc^2 y - 1$ in the final line. Then, since $x = \csc y$ we have,

$\begin{align*} &&(D(\operatorname{arccsc} x))^2 &= \frac{1}{x^2 (x^2 - 1)} \\ \implies && D (\arcsec x) &= \frac{-1}{|x|\sqrt{x^2-1}}. \qquad \blacksquare \end{align*}$

Establish the formula for the derivative of arcsec x

by RoRi

November 29, 2015

Establish the following formula for the derivative of $\arcsec x$ is correct,

$D(\operatorname{arcsec} x) = \frac{1}{|x|\sqrt{x^2-1}}, \qquad \text{for all } |x| >1.$

For $x > 1$ let

$y = \operatorname{arcsec} x \quad \implies \quad x = \sec y.$

Then we know

$\frac{dx}{dy} = \sec y \tan y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && \sec y \tan y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{1}{\sec y \tan y} \\ && \implies && D(\operatorname{arcsec} x) &= \frac{1}{\sec y \tan y} \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \left( \frac{1}{\sec y \tan y} \right)^2 \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \frac{1}{\sec^2 y \tan^2 y} \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \frac{1}{\sec^2 y (\sec^2 y - 1)}. \end{align*}$

Where we use the trig identity $\tan^2 y = \sec^2 y - 1$ in the final line. Then, since $x = \sec y$ we have,

$\begin{align*} &&(D(\operatorname{arcsec} x))^2 &= \frac{1}{x^2 (x^2 - 1)} \\ \implies && D (\operatorname{arcsec} x) &= \frac{1}{|x|\sqrt{x^2-1}}. \qquad \blacksquare \end{align*}$

Establish the formula for the derivative of arccot x

by RoRi

November 23, 2015

Establish the following formula for the derivative of $\arctan x$ is correct,

$D(\operatorname{arccot} x) = \frac{-1}{1+x^2}, \qquad \text{for all } x \in \mathbb{R}.$

For $x \in \mathbb{R}$ let

$y = \operatorname{arccot} x \quad \implies \quad x = \cot y.$

Then we know

$\frac{dx}{dy} = -\csc^2 y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\csc^2 y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\csc^2 y} \\ && \implies && D(\operatorname{arccot} x) &= \frac{-1}{\csc^2 y}. \end{align*}$

Using a trig identity for tangent and cosecant we have

$\csc^2 y = 1 + \cot^2 y = 1 + x^2$

since $x = \cot y$ . Therefore we conclude,

$D(\operatorname{arccot} x) = \frac{-1}{1+x^2}.$

Establish the formula for the derivative of arctan x

by RoRi

November 23, 2015

Establish the following formula for the derivative of $\arctan x$ is correct,

$D(\arctan x) = \frac{1}{1+x^2}, \qquad \text{for all } x \in \mathbb{R}.$

For $x \in \mathbb{R}$ let

$y = \arctan x \quad \implies \quad x = \tan y.$

Then we know

$\frac{dx}{dy} = \sec^2 y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && \sec^2 y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{1}{\sec^2 y} \\ && \implies && D(\arctan x) &= \frac{1}{\sec^2 y}. \end{align*}$

Using a trig identity for tangent and secant we have

$\sec^2 y = 1 + \tan^2 y = 1 + x^2$

since $x = \tan y$ . Therefore we conclude,

$D(\arctan x) = \frac{1}{1+x^2}.$

Establish the formula for the derivative of arccos x

by RoRi

November 23, 2015

Establish the following formula for the derivative of $\arccos x$ is correct,

$D(\arccos x) = \frac{-1}{\sqrt{1-x^2}}, \qquad \text{if } -1<x<1.$

For $-1<x<1$ let

$y = \arccos x \quad \implies \quad x = \cos y.$

Then we know

$\frac{dx}{dy} = -\sin y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\sin y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\sin y} \\ && \implies && D(\arccos x) &= \frac{-1}{\sin y}. \end{align*}$

Using the pythagorean identity for sine and cosine we have

$\begin{align*} \sin^2 y + \cos^2 y =1 && \implies && \sin y &= \pm \sqrt{1-\cos^2 y} \\ && \implies && \sin y &= \pm \sqrt{1-x^2} \end{align*}$

since $x = \cos y$ . Furthermore, since $\sin y \geq 0$ on the range of $\arccos x$ (i.e., $\sin y \ge 0$ for $y \in [0, \pi]$ ) we must take the positive square root. Therefore we conclude,

$D(\arccos x) = \frac{-1}{\sqrt{1-x^2}}.$

Show that a function is monotonic and find a formula for its inverse

by RoRi

September 5, 2015

Let

$f(x) = \begin{cases} x & \text{if } x < 1, \\ x^2 & \text{if } 1 \leq x \leq 4, \\ 8x^{\frac{1}{2}} & \text{if } x > 4. \end{cases}$

Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is strictly increasing on $\mathbb{R}$ we note that it is strictly increasing on each component (since $x^2, \ x,$ and $8x^{1/2}$ are all increasing functions on the domains given). Then we must consider the transition from one of these regions to another. (In other words, we know the function is increasing on each interval, but we need to check that it is increasing from one interval to the next.)

$x_1 < 1 \leq x_2 \quad \implies \quad x_1 < x_2^2 \quad \implies \quad f(x_1) < f(x_2),$

and,

$1 \leq x_1 \leq 4 < x_2 \quad \implies \quad x_1^2 < 8x_2^{\frac{1}{2}} \quad \implies \quad f(x_1) < f(x_2).$

Thus, $f$ is indeed increasing on all of $\mathbb{R}$ .
Next,

$g(y) = \begin{cases} y & \text{if } y < 1, \\ y^{\frac{1}{2}} & \text{if } 1 \leq y \leq 16, \\ \left( \frac{y}{8}\right)^2 & \text{if } y > 16. \end{cases}$

Show a function is monotonic and find a formala for its inverse

by RoRi

September 5, 2015

Let $f(x) = x^3$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then we consider

$\begin{align*} x_2^3 - x_1^3 &= (x_2 - x_1)(x_2^2 + x_1x_2 + x_1^2)\\ &= (x_2 - x_1)\left( \left( x_2 + \frac{x_1}{2} \right)^2 + \frac{3x_1^2}{4} \right). \end{align*}$

But, since $x_2 > x_1$ by assumption, we have $(x_2 - x_1) > 0$ . The second term in the product is also positive since it is a sum of positive terms. Therefore,

$x_2^3 - x_1^3 > 0 \qquad \implies \qquad f(x_2) > f(x_1).$

Hence, $f$ is strictly increasing on $\mathbb{R}$ . (Of course, there are faster ways to discover the $f(x) =x^3$ is increasing, but we do not know yet what is a derivative.)

Next,

$y = x^3 \quad \implies \quad x = y^{\frac{1}{3}} \quad \implies \quad g(y) = y^{\frac{1}{3}} \quad \text{for all } y \in \mathbb{R}.$

Show a function is monotonic and find a formula for its inverse

by RoRi

September 5, 2015

Let $f(x) = 1-x$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then

$x_1 < x_2 \quad \implies \quad -x_1 > -x_2 \quad \implies \quad 1-x_1 > 1-x_2 \quad \implies \quad f(x_1) > f(x_2).$

Hence, $f$ is strictly decreasing on $\mathbb{R}$ .

Next,

$\begin{align*} y = 1-x && \implies && x &= 1-y \\ && \implies && g(y) &= 1-y \qquad \text{for all } y \in \mathbb{R}. \end{align*}$