Chapter 3, Problem 16P

(a)

To determine

To find:

The $x$ and $y$ -intercept and domain or range of the given cubic function.

Answer to Problem 16P

  $\text{points}=\left(0,0\right)\left(2,0\right)\left(6,0\right)$

Explanation of Solution

Given:

The cubic function is

  $f\left(x\right)=x\left(x-2\right)\left(x-6\right)$

Concept used:

The slope of the tangent to a curve $f\left(x\right)$ at any point a is given by the value of the first derivative of the slope at the point $x=a$ .

The tangent to be horizontal so the slope should be equal to 0

That is $\frac{dy}{dx}=0$ .

Calculation:

The cubic function is

  $f\left(x\right)=x\left(x-2\right)\left(x-6\right)\mathrm{................................}\left(1\right)$

  $x-$ interceptat the given function

  $\begin{array}{l}y=0\\ x\left(x-2\right)\left(x-6\right)=0\\ x=0,2,6\end{array}$

  $y-$ interceptat the given function

  $\begin{array}{l}x=0\\ 0\left(0-2\right)\left(0-6\right)\\ 0\end{array}$

The domain of the function $f\left(x\right)$ is the set of all the real numbers.

The leading coefficient is positive

So, the graph of $f\left(x\right)$ is up on the right side and down on the left sideand the graph cut the $x-\text{axis}\text{at}\text{x=0,2,6}$ .

  $\begin{array}{l}f\left(x\right)=x\left(x-2\right)\left(x-6\right)\\ \text{points}=\left(0,0\right)\left(2,0\right)\left(6,0\right)\end{array}$

(b)

To determine

To find:

The first derivative of function for the relative minimum and maximum points.

Answer to Problem 16P

  $f\left(c\right)=0$

Explanation of Solution

Given:

The cubic function is

  $f\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)$

Concept used:

The slope of the tangent to a curve $f\left(x\right)$ at any point a is given by the value of the first derivative of the slope at the point $x=a$ .

The tangent to be horizontal so the slope should be equal to 0

That is $\frac{dy}{dx}=0$ .

Calculation:

The cubic function is

  $\begin{array}{l}f\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)\\ f\left(x\right)=x^3-\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x-abc\mathrm{..................}\left(1\right)\end{array}$

Differentiating equation $\left(1\right)$ with respect to $x$ axis

  $\frac{dy}{dx}=3x^2-2\left(a+b+c\right)x+\left(ab+bc+ca\right)$

Again Differentiating equation $\left(1\right)$ with respect to $x$ axis

  $\frac{d^{2}y}{d{x}^2}=6x-2\left(a+b+c\right)$

For maximum and minimum

  $\begin{array}{l}f\left(c\right)=\left(c-a\right)\left(c-b\right)\left(c-c\right)\\ f\left(c\right)=0\end{array}$