Chapter 2.7, Problem 20E

To determine

To calculate: The first and second derivative of the function $g\left(z\right)=z+\frac{1}{z}$ for $z>0$ also sketch the graph of the function.

Answer to Problem 20E

The value of first derivative of the function is $g\text{'}\left(z\right)=1-\frac{1}{z^2}$ and second derivative is $g\text{'}\text{'}\left(z\right)=\frac{2}{z^3}$ . The graph of the function $g\left(z\right)=z+\frac{1}{z}$ for $z>0$ is,

  

Explanation of Solution

Given information:

The function $g\left(z\right)=z+\frac{1}{z}$ .

Formula used:

Let a function g be continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

The point where the graph changes it nature is known as the point of inflection.

Power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ ,

Calculation:

Consider the provided function $g\left(z\right)=z+\frac{1}{z}$ .

Rewrite the function with help negative exponent that is,

  $\begin{array}{c}g\left(z\right)=z+\frac{1}{z}\\ =z+z^{-1}\end{array}$

Evaluate the first derivative of the function.

Apply sum rule of differentiation,

  $\begin{array}{c}\frac{d}{dz}g\left(z\right)=\frac{d}{dz}\left(z+z^{-1}\right)\\ =\frac{d}{dz}\left(z\right)+\frac{d}{dz}\left(z^{-1}\right)\end{array}$

Apply the power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{d}{dz}g\left(z\right)=\frac{d}{dz}\left(z\right)+\frac{d}{dz}\left(z^{-1}\right)\\ =1-1\cdot z^{-1-1}\\ =1-z^{-2}\\ =1-\frac{1}{z^2}\end{array}$

Evaluate the second derivative of the function, differentiate the first derivative again with respect to z .

  $\frac{d^2}{d{z}^2}g\left(z\right)=\frac{d}{dz}\left(1-z^{-2}\right)$

Apply sum rule of differentiation,

  $\begin{array}{c}\frac{d^2}{d{z}^2}g\left(z\right)=\frac{d}{dz}\left(1-z^{-2}\right)\\ =\frac{d}{dz}\left(1\right)+\frac{d}{dz}\left(-z^{-2}\right)\end{array}$

Apply the power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{d^2}{d{z}^2}g\left(z\right)=\frac{d}{dz}\left(1\right)+\frac{d}{dz}\left(-z^{-2}\right)\\ =0-\left(-2\right)z^{-2-1}\\ =2z^{-3}\\ =\frac{2}{z^3}\end{array}$

Recall if first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

To sketch the graph of the function $f\left(x\right)=\frac{1}{x^2}$ follow the steps below,

Observe that first derivative of the function $g\text{'}\left(z\right)=1-\frac{1}{z^2}$ .

Equate $g\text{'}\left(z\right)=0$ and solve for z ,

  $\begin{array}{c}1-\frac{1}{z^2}=0\\ 1=\frac{1}{z^2}\\ z^2=1\\ z=\pm 1\end{array}$

When $0<z<1$ , $g\left(z\right)=z+\frac{1}{z}$ is a decreasing function.

When $z>1$ , $g\left(z\right)=z+\frac{1}{z}$ is an increasing function.

So, $z=1$ is a point of inflexion as the graph changes it nature from decreasing to increasing.

Next observe that second derivative of the function $g\text{'}\text{'}\left(z\right)=\frac{2}{z^3}$ is positive for $z>0$ , so $g\left(z\right)=z+\frac{1}{z}$ is always concave up.

Therefore, the graph of the function $g\left(z\right)=z+\frac{1}{z}$ for $z>0$ is provided below,

  

Thus, the value of first derivative of the function is $g\text{'}\left(z\right)=1-\frac{1}{z^2}$ and second derivative is $g\text{'}\text{'}\left(z\right)=\frac{2}{z^3}$ .