Chapter 2.7, Problem 21E

To determine

To calculate: The first and second derivative of the function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ also sketch the graph of the function.

Answer to Problem 21E

The value of first derivative of the function is $h\text{'}\left(x\right)=-3x^2+12x-11$ and second derivative is $h\text{'}\text{'}\left(x\right)=-6x+12$ . The graph of the function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ is,

  

Explanation of Solution

Given information:

The function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ .

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 $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ .

Simplify the function, multiple the terms of second and third bracket first,

  $\begin{array}{c}h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)\\ =\left(1-x\right)\left(6-2x-3x+x^2\right)\\ =\left(1-x\right)\left(6-5x+x^2\right)\end{array}$

Now, multiply the terms of both the brackets together,

  $\begin{array}{c}h\left(x\right)=1\left(6-5x+x^2\right)-x\left(6-5x+x^2\right)\\ =6-5x+x^2-6x+5x^2-x^3\\ =6-11x+6x^2-x^3\end{array}$

Evaluate the first derivative of the function,

Apply sum rule of differentiation,

  $\begin{array}{c}\frac{d}{dx}h\left(x\right)=\frac{d}{dx}\left(6-11x+6x^2-x^3\right)\\ =\frac{d}{dx}\left(6\right)-\frac{d}{dx}\left(11x\right)+\frac{d}{dx}\left(6x^2\right)-\frac{d}{dx}\left(x^3\right)\end{array}$

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

  $\begin{array}{c}\frac{d}{dx}h\left(x\right)=-11+12x-3x^2\\ =-3x^2+12x-11\end{array}$

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

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

  $\begin{array}{c}\frac{d^2}{d{x}^2}h\left(x\right)=\frac{d}{dx}\left(-3x^2+12x-11\right)\\ =\frac{d}{dx}\left(-3x^2\right)+\frac{d}{dx}\left(12x\right)-\frac{d}{dx}\left(11\right)\\ =-6x+12\end{array}$

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 $\left[c,d\right]$ .

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

To sketch the graph of the function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ follow the steps below,

Observe that first derivative of the function $h\text{'}\left(x\right)=-3x^2+12x-11$ .

The derivative of the function has solutions,

  $\begin{array}{c}-3x^2+12x-11=0\\ 3x^2-12x+11=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{12\pm \sqrt{144-132}}{6}\end{array}$

Simplify it further as,

  $\begin{array}{c}x=\frac{12\pm \sqrt{12}}{6}\\ x=\frac{12\pm 2\sqrt{3}}{6}\\ x=\frac{6\pm \sqrt{3}}{3}\\ x=2.577,1.422\end{array}$

Next observe that second derivative of the function $h\text{'}\text{'}\left(x\right)=-6x+12$ is positive when,

  $\begin{array}{c}-6x+12\ge 0\\ 12\ge 6x\\ 2\ge x\\ x\le 2\end{array}$

The function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ is concave up.

Next observe that second derivative of the function $h\text{'}\text{'}\left(x\right)=-6x+12$ is negative when,

  $\begin{array}{c}-6x+12<0\\ 12<6x\\ 2<x\end{array}$

The function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ is concave down.

At $x=2$ , the function changes it nature so it is a point of inflection.

Therefore, the graph of the function $h\left(x\right)=\left(1-x\right)\left(2-x\right)\left(3-x\right)$ is provided below,

  

Thus, the value of first derivative of the function is $h\text{'}\left(x\right)=-3x^2+12x-11$ and second derivative is $h\text{'}\text{'}\left(x\right)=-6x+12$ .