Chapter 5, Problem 2P

a.

To determine

To graph: several members of the family of functions $f(x)=\frac{(2cx-x^{2)}}{c^3}\text{for}c>0$ and look at the region enclosed by these curves and the x −axis and make a conjecture about how the areas of these regions are related.

Answer to Problem 2P

Explanation of Solution

Given information: Given several members of the family of functions $f(x)=\frac{(2cx-x^{2)}}{c^3}\text{for}c>0$

Calculation:

Here’s a sketch for c =1 (green), c = 2 (orange),), c = 3 (purple) is given below.

  

Notice how the maximum width of the region is always equal to 2c and the height to $\frac{1}{c}$ . This seems to indicate that the region remains constant for any positive value of c .

b.

To determine

To prove: conjecture in part (a).

Answer to Problem 2P

Explanation of Solution

Given information:

Calculation:

Here’s a sketch for c =1 (green), c = 2 (orange),), c = 3 (purple) is given below.

  

First need to prove conjecture that the width of the region is always 2c .

  $\begin{array}{c}\frac{2cx-x^2}{c^3}=0\to 2cx-x^2=0\\ x=0,\text{otherwise}\\ 2c=x\end{array}$

The area of the region by integrating the function with respect to x between 0 and 2c.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{2c}{\int }}\frac{2cx-x^2}{c^3}}dx=\frac{1}{c^3}{\displaystyle \underset{0}{\overset{2c}{\int }}[2cx-x^2}]dx\\ =\frac{1}{c^3}\left[c{x}^2-\frac{1}{3}{x}^3\right]\begin{array}{c}2c\\ 0\end{array}\\ =\frac{1}{c^3}\left[(4c^3-\frac{8}{3}{c}^3)-(0)\right]\\ =\frac{4}{3}\end{array}$

Notice how c disappears from the equation and the region remains constant, as had first conjectured.

c.

To determine

To sketch: the curve traced out by the vertices (highest points) of the family of functions and finds what kind of curve this is.

Answer to Problem 2P

Explanation of Solution

Given information:

Calculation:

Here’s a sketch for c =1 (green), c = 2 (orange),), c = 3 (purple) is given below.

  

Drawing a line over the vertices, notice that it converges towards zero as x increases, and its derivative, always negative, also converges to zero as x increases and also notice that the vertices are $\frac{1}{c}$ away from the x −axis, which suggests the black line is the graph of a function with x in the denominator. But can prove that.

d.

To determine

To find: an equation of the curve sketched in part (c).

Answer to Problem 2P

  $f(x)=\frac{1}{x},x>0$ .

Explanation of Solution

Given information:

Calculation:

Here’s a sketch for c =1 (green), c = 2 (orange),), c = 3 (purple) is given below.

  

  $\begin{array}{c}f\text{'}(x)=\frac{d}{dx}\left[\frac{2cx-x^2}{c^3}\right]=\frac{2c-2x}{c^3}\\ \frac{2c-2x}{c^3}=0\to x=c\\ f(c)=\frac{2c^2-c^2}{c^3}=\frac{1}{c}\end{array}$

The y -coordinating of the vertex by setting x = c in the original function. Hence, the vertex of $f(x)=\frac{(2cx-x^2)}{c^3}$ , for any c >0 will always have coordinates ( c , $\frac{1}{c}$ ). The function seeks returns:

  $f(x)=\frac{1}{x},x>0.$