Chapter 3.3, Problem 39E

To determine

To calculate: The points on the curve where the tangent is parallel to the x -axis.

Answer to Problem 39E

The required points are $\left(2,0\right)$ and $\left(-1,27\right)$ .

Explanation of Solution

Given information:

The curve: $y=2x^3-3x^2-12x+20$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Point-slope form of a line:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The given curve is

  $y=2x^3-3x^2-12x+20$  .......... (1)

To find the points on the curve where the tangent is parallel to the curve, differentiate the given curve.

Use the power rule of differentiation $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(2x^3\right)-\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(12x\right)+\frac{d}{dx}\left(20\right)\\ =6x^2-6x-12+0\\ =6x^2-6x-12\end{array}$

Now, set the above derivative equal to 0 and simplify for x .

  $\begin{array}{c}6x^2-6x-12=0\\ x^2-x-2=0\\ x^2-2x+x-2=0\\ x\left(x-2\right)+1\left(x-2\right)=0\end{array}$

Simplify further.

  $\begin{array}{c}x\left(x-2\right)+1\left(x-2\right)=0\\ \left(x-2\right)\left(x+1\right)=0\end{array}$

Solve for x .

  $\begin{array}{c}x-2=0\\ x=2\end{array}$

And,

  $\begin{array}{c}x+1=0\\ x=-1\end{array}$

So, the values of x are $-1$ and 2.

Substitute $-1$ for x in the curve (1) and simplify.

  $\begin{array}{c}y=2{\left(-1\right)}^3-3{\left(-1\right)}^2-12\left(-1\right)+20\\ =-2-3+12+20\\ =-5+32\\ =27\end{array}$

Substitute 2 for x in the curve (1) and simplify.

  $\begin{array}{c}y=2{\left(2\right)}^3-3{\left(2\right)}^2-12\left(2\right)+20\\ =16-12-24+20\\ =4-4\\ =0\end{array}$

Thus, for $x=-1$ ; $y=27$

And for $x=2$ ; $y=0$

Hence, the required points are $\left(2,0\right)$ and $\left(-1,27\right)$ .