Chapter 9, Problem 1RE

(A)

To determine

To find: The change in $y=2x^2+5$ if x changes from 1 to 3.

Answer to Problem 1RE

The change in y is 16.

Explanation of Solution

Given:

The function y is $f\left(x\right)=2x^2+5$.

Definition used:

Increments:

“For $y=f\left(x\right)$, $\Delta x=x_2-x_1$ and $\Delta y=y_2-y_1$”.

Calculation:

Consider the function $f\left(x\right)=2x^2+5$.

Compute the change in x or $\Delta x$ for the function $f\left(x\right)=2x^2+5$ for $x_1=1\text{ and }{x}_2=3$.

From the given definition, it is known that $\Delta x=x_2-x_1$.

$\begin{array}{c}\Delta x=x_2-x_1\\ =3-1\\ =2\end{array}$

So the value of change in x is $\Delta x=2$.

Compute the change in y or $\Delta y$ for the function $f\left(x\right)=2x^2+5$ for $x_1=1\text{ and }{x}_2=4$.

Compute $f\left(1\right)$ by substituting $x=2$ in the function $f\left(x\right)=2x^2+5$.

$\begin{array}{c}f\left(1\right)=2{\left(1\right)}^2+5\\ =2+5\\ =7\end{array}$

Therefore, the value of $f\left(1\right)$ is 7.

Compute $f\left(3\right)$ by substituting $x=3$ in the function $f\left(x\right)=2x^2+5$.

$\begin{array}{c}f\left(3\right)=2{\left(3\right)}^2+5\\ =2\times 9+5\\ =23\end{array}$

Therefore, the value of $f\left(3\right)$ is 23.

From the given definition, it is known that $\Delta y=y_2-y_1$.

$\begin{array}{c}\Delta y=f\left(x_2\right)-f\left(x_1\right)\\ =f\left(3\right)-f\left(1\right)\\ =23-7\\ =16\end{array}$

Therefore, the change in y as x changes from 1 to 3 is 16.

(B)

To determine

To find: The average rate of change of y with respect to x when the value of x changes from 1 to 3.

Answer to Problem 1RE

The average rate of change of y with respect to x is 8.

Explanation of Solution

Definition used:

Average Rate of Change:

“For $y=f\left(x\right)$, the average rate of change form $x=a$ to $x=a+h$ is given by

$\frac{f\left(a+h\right)-f\left(a\right)}{\left(a+h\right)-a}=\frac{f\left(a+h\right)-f\left(a\right)}{h}h\ne 0$”.

Calculation:

From part (A), the value of $f\left(3\right)$ is 23 and the value of $f\left(1\right)$ is 7.

Compute the average rate of change by using the above mentioned definition.

$\begin{array}{c}\text{Average rate of change}=\frac{f\left(1+2\right)-f\left(1\right)}{\left(1+2\right)-1}\\ =\frac{f\left(3\right)-f\left(1\right)}{3-1}\\ =\frac{23-7}{2}\\ =\frac{16}{2}\end{array}$

$=8$

Therefore, the average rate of change of y with respect to x when the value of x changes from 1 to 3 is 8.

(C)

To determine

To find: The slope of the secant line passing through the points $\left(1,f\left(1\right)\right)$ and $\left(3,f\left(3\right)\right)$.

Answer to Problem 1RE

The slope of the secant line is 8.

Explanation of Solution

Result used:

“The slope of the secant line joining points $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right)$ is $\frac{f\left(a\right)-f\left(b\right)}{a-b}$”.

Calculation:

From part (A), the value of $f\left(3\right)$ is 23 and the value of $f\left(1\right)$ is 7.

Compute the slope of the secant line joining points $\left(1,f\left(1\right)\right)$ and $\left(3,f\left(3\right)\right)$ as follows,

$\begin{array}{c}\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{23-7}{2}\\ =\frac{16}{2}\\ =8\end{array}$

Therefore, the slope of the secant line passing through the points $\left(1,f\left(1\right)\right)$ and $\left(3,f\left(3\right)\right)$ is 8.

(D)

To determine

To find: The instantaneous rate of change of y with respect to x when $x=1$.

Answer to Problem 1RE

The instantaneous rate of change of y is 4.

Explanation of Solution

Definition used:

Instantaneous rate of change:

“For $y=f\left(x\right)$, the instantaneous rate of change at $x=a$ is $\underset{h\to 0}{\lim }\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}\right)$ if limit exists”.

Calculation:

Obtain instantaneous rate of change at $x=1$.

That is, $\underset{h\to 0}{\lim }\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}\right)=\underset{h\to 0}{\lim }\left(\frac{f\left(1+h\right)-f\left(1\right)}{h}\right)$ (1)

Obtain $f\left(1+h\right)$.

$\begin{array}{c}f\left(1+h\right)=2{\left(1+h\right)}^2+5\\ =2\left(1+2h+h^2\right)+5\\ =2+4h+2h^2+5\\ =7+4h+2h^2\end{array}$

From part (A), $f\left(1\right)=7$.

Substitute the value of $f\left(1+h\right),f\left(1\right)$ in the equation (1),

$\begin{array}{c}\underset{h\to 0}{\lim }\left(\frac{f\left(1+h\right)-f\left(1\right)}{h}\right)=\underset{h\to 0}{\lim }\left(\frac{f\left(1+h\right)}{h}-\frac{f\left(1\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{7+4h+2h^2}{h}-\frac{7}{h}\right)\\ =\underset{h\to 0}{\lim }\left(4+2h\right)\\ =4\end{array}$

Therefore, the instantaneous rate of change of y with respect to x at $x=1$ is 4.

(E)

To determine

To find: The slope of the tangent line at $x=1$.

Answer to Problem 1RE

The slope of the tangent line is 4.

Explanation of Solution

Formula used:

Slope of the tangent line.

The slope of the tangent line at the point $x=a$ is the value of the derivative of the function at $x=a$.

Calculation:

From part (D), the instantaneous rate of change of y with respect to x at $x=1$ is 4.

That is, the value of the derivative of the function $f\left(x\right)=2x^2+5$ at $x=1$ is 4.

So by the above mentioned definition, the slope of the tangent line is 4.

Therefore, the slope of the tangent line at $x=1$ is 4.

(F)

To determine

To find: The value of $f^{\prime }\left(1\right)$.

Answer to Problem 1RE

The value of $f^{\prime }\left(1\right)$ is 4.

Explanation of Solution

Theorem used:

Power rule:

“If $f\left(x\right)=x^n$, where n is a real number, then $f^{\prime }\left(x\right)=n{x}^{\left(n-1\right)}$.

Also, $y^{\prime }=n{x}^{\left(n-1\right)}$ and $\frac{dy}{dx}=n{x}^{\left(n-1\right)}$”.

Sum and difference property:

“If $f\left(x\right)=u\left(x\right)\pm v\left(x\right)$, then $f^{\prime }\left(x\right)=u^{\prime }\left(x\right)\pm v^{\prime }\left(x\right)$.

Also, $y^{\prime }=u^{\prime }\pm v^{\prime }$ and $\frac{dy}{dx}=\frac{du}{dx}\pm \frac{dv}{dx}$”.

Constant Function rule:

“If $f\left(x\right)=C$ is a constant function, then $f^{\prime }\left(x\right)=0$.

Also, $y^{\prime }=0$ and $\frac{dy}{dx}=0$”.

Calculation:

Compute the derivative of $f\left(x\right)$.

$\begin{array}{c}{f}^{\prime }\left(x\right)={\left(2x^2+5\right)}^{\prime }\\ ={\left(2x^2\right)}^{\prime }+{\left(5\right)}^{\prime }\left[\text{By sum and difference rule}\right]\\ =4x+0\left[\text{By power and constant rule}\right]\\ =4\end{array}$

Substitute $x=1$ in $f^{\prime }\left(x\right)=4x$ and simplify it as follows.

$\begin{array}{c}{f}^{\prime }\left(x\right)=4\left(1\right)\\ =4\end{array}$

Therefore, the value of $f^{\prime }\left(1\right)$ is 4.