Chapter 3, Problem 2P

To determine

To find: The point of the tangent line and sketch the graph of the curve and tangent line.

Answer to Problem 2P

The equation of the tangent line to the curve at $\left(2,6\right)$ is, $y=9x-12$.

Explanation of Solution

Given:

The curves is $y=x^3-3x+4\text{ and }y=3\left(x^2-x\right)$.

Formula used:

The equation of the tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$ (1)

Where, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1,y=y_1}$

Calculation:

Obtain the derivative of $y_1=x^3-3x+4$.

$\begin{array}{c}\frac{d{y}_1}{dx}=\frac{d}{dx}\left(x^3-3x+4\right)\\ =3x^{3-1}-3x^{1-1}+0\\ =3x^2-3\end{array}$

Obtain the derivative of $y_2=3\left(x^2-x\right)$.

$\begin{array}{c}\frac{d{y}_2}{dx}=\frac{d}{dx}\left(3\left(x^2-x\right)\right)\\ =3\left(2x^{2-1}-1x^{1-1}\right)\\ =6x-3\end{array}$

Since the slopes of the tangent of the two curves are equal. That is $\frac{d{y}_1}{dx}=\frac{d{y}_2}{dx}$.

$\begin{array}{l}3x^2-3=6x-3\\ 3x^2-6x=0\\ 3x\left(x-2\right)=0\\ x=0\text{ or }x=2\end{array}$

Thus, the derivative of the equation is $\frac{d{y}_1}{dx}=3x^2-3\text{ and }\frac{d{y}_2}{dx}=6x-3$.

The slope of tangent to the curves at $x=0$ in $\frac{d{y}_1}{dx}=3x^2-3$.

$\begin{array}{c}\frac{d{y}_1}{dx}=3{\left(0\right)}^2-3\\ =0-3\\ =-3\end{array}$

The slope of the tangent is −3 but curve do not intersect.

The slope of tangent to the curves at $x=2$ in $\frac{d{y}_2}{dx}=6x-3$.

$\begin{array}{c}\frac{d{y}_1}{dx}=6\left(2\right)-3\\ =12-3\\ =9\end{array}$

The slope of the tangent is $m=9$ and the curve intersect at point $\left(2,6\right)$.

Thus, the common tangent line at $\left(2,6\right)$.

Substitute $\left(2,6\right)$ for $\left(x_1,y_1\right)$ and slope $m=9$ in equation (1),

$\begin{array}{l}y-6=9\left(x-2\right)\\ y-6=9x-18\\ y=9x-18+6\\ y=9x-12\end{array}$

Thus, the equation of the tangent line to the curve at $\left(2,6\right)$ is $y=9x-12$.

Therefore, the required point is, $\left(2,6\right)$.

Use the online graphing calculator and draw the given curves and obtained tangent line as shown in Figure 1.

From Figure 1, it is observed that the given curves $y=x^3-3x+4\text{ and }y=3\left(x^2-x\right)$ are tangent to each other at the point $\left(2,6\right)$.