Chapter 2.7, Problem 45E

To determine

To find: The first and second derivative of the function $f\left(x\right)=3x^2+2x+1$.

Answer to Problem 45E

The first and second derivatives of the function are, $f^{\prime }\left(x\right)=6x+2$ and $f^{″}\left(x\right)=6$.

Explanation of Solution

Formula used:

The derivative of a function f , denoted by $f^{\prime }\left(x\right)$, is

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (1)

Calculation:

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

Compute $f^{\prime }\left(x\right)$ by using the equation (1),

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{3{\left(x+h\right)}^2+2\left(x+h\right)+1-\left\{3x^2+2x+1\right\}}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(3x^2+3h^2+6xh\right)+\left(2x+2h\right)+1-\left\{3x^2+2x+1\right\}}{h}\\ =\underset{h\to 0}{\lim }\frac{3x^2+3h^2+6xh+2x+2h+1-3x^2-2x-1}{h}\end{array}$

Simplify the numerator and to obtain the value of $f^{\prime }\left(x\right)$,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{3h^2+6xh+2h}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(3h+6x+2\right)}{h}\end{array}$

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\left(3h+6x+2\right)\\ =3\left(0\right)+6x+2\\ =6x+2\end{array}$

Thus, the first derivative of the function is, $f^{\prime }\left(x\right)=6x+2$.

Obtain the second derivative of the function.

Compute $f^{″}\left(x\right)$ by using the equation (1),

$\begin{array}{c}{f^{\prime }}^{\prime }\left(x\right)={\left(f^{\prime }\left(x\right)\right)}^{\prime }\\ =\underset{h\to 0}{\lim }\frac{f^{\prime }\left(x+h\right)-f^{\prime }\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{6\left(x+h\right)+2-\left\{6x+2\right\}}{h}\\ =\underset{h\to 0}{\lim }\frac{6x+6h+2-6x-2}{h}\end{array}$

Simplify the numerator and to obtain the derivatives $f^{″}\left(x\right)$,

$f^{″}\left(x\right)=\underset{h\to 0}{\lim }\frac{6h}{h}$

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

$\begin{array}{c}{f}^{″}\left(x\right)=\underset{h\to 0}{\lim }\left(6\right)\\ =6\end{array}$

Thus, the second derivative of the function is, $f^{″}\left(x\right)=6$.

To check: The derivatives $f\left(x\right)$, $f^{\prime }\left(x\right)$,.and $f^{″}\left(x\right)$ are reasonable by comparing the graphs of $f\left(x\right)$, $f^{\prime }\left(x\right)$. and $f^{″}\left(x\right)$.

The derivative $f\left(x\right)$, $f^{\prime }\left(x\right)$. and $f^{″}\left(x\right)$ are reasonable.

Graph:

Use the online graphing calculator to draw the graph of the functions $f\left(x\right),f^{\prime }\left(x\right)\text{ and }{f}^{″}\left(x\right)$ as shown below in Figure 1.

From Figure 1, it is observed that the graph of $f^{\prime }\left(x\right)$ is a straight line and the graph of $f^{″}\left(x\right)$ is a horizontal line.

Take several points on the domain and estimate the slope of the tangent to the function $f\left(x\right)$ which is same as the point of $f^{\prime }\left(x\right)$ at that point.

Thus, the derivative $f^{\prime }\left(x\right)$ is reasonable.

The $f^{\prime }\left(x\right)$ represents a linear function and $f^{″}\left(x\right)$ represents a constant function.

The straight line has same slope at all points. That is, $f^{″}\left(x\right)$ is constant value.

From the Figure 1, the estimated slope of the function $f^{\prime }\left(x\right)$ is 6 which is same as the derivative of $f^{\prime }\left(x\right)$. That is, $f^{″}\left(x\right)=6$.

Therefore, the function $f^{″}\left(x\right)$ is reasonable.