## Precalculus (6th Edition) Blitzer

a) The average rate of change in the area is $12.1\text{ square inches}$ per inch as x changes from $6$ to $6.1\text{ inches}$ and $12.01\text{ square inches}$ per inch as x changes from $6$ to $6.01\text{ inches}$. b) The instantaneous rate of change of the area with respect to x is $12\text{ square inches}$ per inch when $x=6\text{ inches}$.
(a) Consider that the area of the square is described by the function $f\left( x \right)={{x}^{2}}$, x changes from $6\text{ inches}$ to $6.1\text{ inches}$ Here, as x changes from $6$ to $6.1$, this implies $a=6$ and $h=0.1$ Now, compute the average rate of change of the area of a square by substituting the above values in the difference quotient as follows: \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 6+0.1 \right)-f\left( 6 \right)}{0.1} \\ & =\frac{f\left( 6.1 \right)-f\left( 6 \right)}{0.1} \end{align} Now, substitute $x=6.1$ and $x=6$ respectively in the function $f\left( x \right)={{x}^{2}}$ and simplify. \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{{{6.1}^{2}}-{{6}^{2}}}{0.1} \\ & =\frac{37.21-36}{0.1} \\ & =12.1 \end{align} Thus, the average rate of change in the area is $12.1$ square inches per inch as x changes from $6$ to $6.1\text{ inches}$. Now, consider that x changes from $6\text{ inches}$ to $6.01\text{ inches}$. Here, as x changes from $6$ to $6.01$, this implies $a=6$ and $h=0.01$ Now, compute the average rate of change of the area of a square by substituting the above values in the difference quotient as follows: \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{f\left( 6+0.01 \right)-f\left( 6 \right)}{0.01} \\ & =\frac{f\left( 6.01 \right)-f\left( 6 \right)}{0.01} \end{align} Now, substitute $x=6.01$ and $x=6$ respectively in the function $f\left( x \right)={{x}^{2}}$ and simplify. \begin{align} & \frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{{{6.01}^{2}}-{{6}^{2}}}{0.01} \\ & =\frac{36.1201-36}{0.01} \\ & =12.01 \end{align} Thus, the average rate of change in the area is $12.01$ square inches per inch as x changes from $6$ to $6.01\text{ inches}$. (b) Consider that the area of the square is described by the function $f\left( x \right)={{x}^{2}}$. Compute the derivative of $f\left( x \right)={{x}^{2}}$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ as follows: To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)={{x}^{2}}$. \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h} \end{align} Now, simplify ${{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}+2xh+{{h}^{2}}-{{x}^{2}}}{h}$ Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2x+h \right)$ Apply the limits, \begin{align} & f'\left( x \right)=2x+0 \\ & =2x \end{align} Now, substitute $x=6$ in $f'\left( x \right)$ to compute the instantaneous change of “f” at $6$. \begin{align} & f'\left( 6 \right)=2\cdot 6 \\ & =12 \end{align} Thus, the instantaneous rate of change of the area with respect to x is $12\text{ square inches}$ per inch when $x=6\text{ inches}$.