Chapter 3.1, Problem 50E

To determine

To find:

The equation of the tangent line to the curve at the given point

Answer to Problem 50E

The tangent is horizontal at $x=-1\pm \frac{1}{3}\sqrt{6}$

Explanation of Solution

Given:

The function is

  $y=x^3+3x^2+x+3$

Concept used:

An equation of a line $l$ through a point $P_1\left(x_1,y_1\right)$ with slope $m$ . If $P\left(x,y\right)$ is any point with $x\ne x_1$

Then P is on $l$ if and only if the slope of the line through $P_1$ and P

The equation is $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=x^3+3x^2+x+\mathrm{3...................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(x^3+3x^2+x+3\right)\\ \frac{dy}{dx}=3x^2+6x+1\end{array}$

The tangent are horizontal

  $\begin{array}{l}\frac{dy}{dx}=0\\ 3x^2+6x+1=0\\ x=\frac{-6\pm \sqrt{6^2-4\left(3\right)\left(1\right)}}{2\left(3\right)}\\ x=\frac{-6\pm \sqrt{36-12}}{6}\\ x=\frac{-6\pm \sqrt{24}}{6}\\ x=-1\pm \frac{1}{3}\sqrt{6}\end{array}$