Chapter 1.4, Problem 4QQ

a.

To determine

To find: $f\left(3\right)$ for $f\left(x\right)=2x-x^2$ .

Answer to Problem 4QQ

  $f\left(3\right)=-3$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=2x-x^2$ .

Concept used:

Replace x in $f\left(x\right)$ by 3 for the value of $f\left(3\right)$ .

Calculation:

On replacing x by 3 in $f\left(x\right)$ .

  $\begin{array}{c}f\left(3\right)=2\left(3\right)-{\left(3\right)}^2\\ =6-9\\ =-3\end{array}$

Hence the value of the function at 3 is $-3$ .

b.

To determine

To find: $f\left(3+h\right)$ for $f\left(x\right)=2x-x^2$ .

Answer to Problem 4QQ

  $f\left(3+h\right)=-h^2-4h-3$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=2x-x^2$ .

Concept used:

Replace x in $f\left(x\right)$ by $3+h$ for the value of $f\left(3+h\right)$ .

Calculation:

On replacing x by $3+h$ in $f\left(3+h\right)$ .

  $\begin{array}{c}f\left(3+h\right)=2\left(3+h\right)-{\left(3+h\right)}^2\\ =6+2h-\left(3^2+h^2+2\left(3h\right)\right)\\ =6+2h-9-h^2-6h\\ =-h^2-4h-3\end{array}$

Hence $f\left(3+h\right)=-h^2-4h-3$ .

c.

To determine

To find: $\frac{f\left(3+h\right)-f\left(3\right)}{h}$ for $f\left(x\right)=2x-x^2$ .

Answer to Problem 4QQ

  $\frac{f\left(3+h\right)-f\left(3\right)}{h}=-h-4$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=2x-x^2$ .

Concept used:

From part (a) and part (b) use the value of $f\left(3\right)$ and $f\left(3+h\right)$ .

Calculation:

  $\begin{array}{c}\frac{f\left(3+h\right)-f\left(3\right)}{h}=\frac{-h^2-4h-3-\left(-3\right)}{h}\\ =\frac{-h^2-4h-3+3}{h}\\ =\frac{-h^2-4h}{h}\\ =\frac{\left(-h-4\right)h}{h}\\ =-h-4\end{array}$

Hence $\frac{f\left(3+h\right)-f\left(3\right)}{h}=-h-4$ .

d.

To determine

To find: the instantaneous rate of change of f at $x=3$ .

Answer to Problem 4QQ

The instantaneous rate of change of f at $x=3$ is $-4$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=2x-x^2$ .

Concept used:

Instantaneous rate of change of any function f with respect to x at any point a is given by: $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

From part c, at $a=3$ , $\frac{f\left(3+h\right)-f\left(3\right)}{h}=-h-4$ .

Apply limit $h\to 0$ on both the sides.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{h}=\underset{h\to 0}{\lim }\left(-h-4\right)\\ =-0-4\\ =-4\end{array}$

Hence $\underset{h\to 0}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{h}=-4$ .

Conclusion:

The instantaneous rate of change of f at $x=3$ is $-4$ .