Chapter 3.3, Problem 8E

To determine

To calculate: The horizontal tangents of the curve.

Answer to Problem 8E

The required horizontal tangents of the curve are $x=\frac{4+\sqrt{13}}{3}$ and $x=\frac{4-\sqrt{13}}{3}$ .

Explanation of Solution

Given information:

The curve: $y=x^3-4x^2+x+2$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Quadratic formula:

The roots of the quadratic equation $a{x}^2+bx+c=0$ can be found by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

The given curve is $y=x^3-4x^2+x+2$ .

Use the power rule on the above curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^3-4x^2+x+2\right)\\ =\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(2\right)\\ =3x^2-8x+1+0\\ =3x^2-8x+1\end{array}$

Now, to find the horizontal tangents of the given curve, set the above derivative equal to zero.

  $3x^2-8x+1=0$

To find the roots of the above expression, use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Substitute $-8$ for $b$ , 3 for a and 1 for c in the above formula and simplify for x .

  $\begin{array}{c}x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(3\right)\left(1\right)}}{2\left(3\right)}\\ =\frac{8\pm \sqrt{64-12}}{6}\\ =\frac{8\pm \sqrt{52}}{6}\\ =\frac{8\pm 4\sqrt{13}}{6}\end{array}$

Simplify the above expression further.

  $\begin{array}{c}x=\frac{8\pm 4\sqrt{13}}{6}\\ =\frac{4\pm \sqrt{13}}{3}\end{array}$

Thus, the roots of the expression are $x=\frac{4\pm \sqrt{13}}{3}$ .

Hence, the required horizontal tangents of the curve are $x=\frac{4+\sqrt{13}}{3}$ and $x=\frac{4-\sqrt{13}}{3}$ .