Chapter 4.5, Problem 8QR

To determine

To complete: The given table where $g\left(x\right)$ is the function whose graph is the tangent line to the graph of $f\left(x\right)=x^3-4x+1$ at $x=1$ .

Answer to Problem 8QR

$x$	$f\left(x\right)$	$g\left(x\right)$
$0.7$	$-1.457$	$-1.7$
$0.8$	$-1.688$	$-1.8$
$0.9$	$-1.871$	$-1.9$
$1$	$-2$	$-2$
$1.1$	$-2.069$	$-2.1$
$1.2$	$-2.072$	$-2.2$
$1.3$	$-2.3$	$-2.3$

Explanation of Solution

Given information: $f\left(x\right)=x^3-4x+1$ at $x=1$ .

Concept used: The slope of the tangent line at $x=x_0$ is the derivative of the function, evaluated at $x=x_0$ , that is, $M\left(x_0\right)=f\text{'}\left(x_0\right)$ . Also, the equation of a tangent line is given as $y-y_0=m\left(x-x_0\right)$ , where $y_0={\left.f\left(x\right)\right|}_{x=c}$ .

Calculation:

Substitute $x=1$ into the given function and simplify it to determine the value of $y_0$ .

  $\begin{array}{c}{y}_0=f\left(1\right)\\ =1^3-4\left(1\right)+1\\ =-2\end{array}$

Differentiate the given function with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x^3-4x+1\right)\\ =\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(4x\right)+\frac{d}{dx}\left(1\right)\\ =3x^2-4\end{array}$

Therefore, $M\left(x_0\right)=f\text{'}\left(x_0\right)=3x_0{}^2-4$ .

Now, substitute $x_0=1$ into the obtained derivative, that is, $f\text{'}\left(x_0\right)=3x_0{}^2-4$ , and simplify it to determine the value of slope m.

  $\begin{array}{c}m=f\text{'}\left(1\right)\\ =3{\left(1\right)}^2-4\\ =-1\end{array}$

It is known that the equation of a tangent line is given as $y-y_0=m\left(x-x_0\right)$ .

Now, substitute $m=-1$ , $x_0=1$ , and $y_0=-2$ into the equation of a tangent line and simplify it.

  $\begin{array}{c}y-\left(-2\right)=-1\left(x-1\right)\\ y+2=-x+1\\ y=-x-1\end{array}$

More simply, the obtained equation of a tangent line can be written as $g\left(x\right)=-x-1$ .

Now, find the values of the function $f\left(x\right)$ and $g\left(x\right)$ at $x=0.7$ , $x=0.8$ , $x=0.9$ , $x=1$ , $x=1.1$ , $x=1.2$ , and $x=1.3$ .

  $\begin{array}{c}f\left(0.7\right)={\left(0.7\right)}^3-4\left(0.7\right)+1\\ =-\text{1}\text{.457}\end{array}$

  $\begin{array}{c}g\left(0.7\right)=-0.7-1\\ =-1.7\end{array}$

  $\begin{array}{c}f\left(0.8\right)={\left(0.8\right)}^3-4\left(0.8\right)+1\\ =-1.688\end{array}$

  $\begin{array}{c}g\left(0.8\right)=-0.8-1\\ =-1.8\end{array}$

  $\begin{array}{c}f\left(0.9\right)={\left(0.9\right)}^3-4\left(0.9\right)+1\\ =-\text{1}\text{.871}\end{array}$

  $\begin{array}{c}g\left(0.9\right)=-0.9-1\\ =-1.9\end{array}$

  $\begin{array}{c}f\left(1\right)={\left(1\right)}^3-4\left(1\right)+1\\ =-\text{2}\end{array}$

  $\begin{array}{c}g\left(1\right)=-1-1\\ =-2\end{array}$

  $\begin{array}{c}f\left(1.1\right)={\left(1.1\right)}^3-4\left(1.1\right)+1\\ =-2.069\end{array}$

  $\begin{array}{c}g\left(1.1\right)=-1.1-1\\ =-2.1\end{array}$

  $\begin{array}{c}f\left(1.2\right)={\left(1.2\right)}^3-4\left(1.2\right)+1\\ =-2.072\end{array}$

  $\begin{array}{c}g\left(1.2\right)=-1.2-1\\ =-2.2\end{array}$

  $\begin{array}{c}f\left(1.3\right)={\left(1.3\right)}^3-4\left(1.3\right)+1\\ =-2.3\end{array}$

  $\begin{array}{c}g\left(1.3\right)=-1.3-1\\ =-2.3\end{array}$

Now, write the obtained values in the given table.

$x$	$f\left(x\right)$	$g\left(x\right)$
$0.7$	$-1.457$	$-1.7$
$0.8$	$-1.688$	$-1.8$
$0.9$	$-1.871$	$-1.9$
$1$	$-2$	$-2$
$1.1$	$-2.069$	$-2.1$
$1.2$	$-2.072$	$-2.2$
$1.3$	$-2.3$	$-2.3$