Chapter 4.4, Problem 23E

To determine

To Find:The vertical and horizontal tangent coordinates of the function.

Answer to Problem 23E

Vertical tangents are

  $\left(\frac{-3}{16},\frac{3}{8}\right),(0,0),(-8,6)$

Horizontal tangents coordinates are $\left(-\frac{2\sqrt{3}-5}{9},-\frac{2\sqrt{3}}{9}\right),\left(\frac{2\sqrt{3}-5}{9},\frac{2\sqrt{3}}{9}\right)$

Explanation of Solution

Given:Use a graph to estimate the coordinates of the leftmost point on the curve $x=t^4-2t^3-2t^2,y=t^3-t$ then find at what point does the curve have the vertical and horizontal tangent.

Given: $x=t^4-2t^3-2t^2,y=t^3-t$

The graph of the function is

  

For vertical tangents

The slope of tangent is at the point $\infty$

Then

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\infty \\ \frac{dx}{dt}=0\\ \frac{d(t^4-2t^3-2t^2)}{dt}=4t^3-2t^3-4=2t(2t^2-3t-2)=0\end{array}$

Factoring we get

  $\begin{array}{l}2t(2t^2-1)=0\\ 2t=0\\ t=0\\ 2t^2-3t-2=0\\ t=-\frac{1}{2},2\\ i.e\\ t\in -\frac{1}{2},0,2\end{array}$

Now for $t=-\frac{1}{2}$

  $(x,y)=\left(-\frac{3}{16},\frac{3}{8}\right)$

For

  $\begin{array}{l}t=0\\ (x,y)=(0,0)\end{array}$

For

  $\begin{array}{l}t=2\\ (x,y)=(-8,6)\end{array}$

Similarly

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=0\\ \frac{dy}{dt}=0\\ \frac{d(t^3-t)}{dt}=3t^2-1=0\\ t=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}$

Solving

  $\begin{array}{l}t=-\frac{1}{\sqrt{3}}\\ x={\left(-\frac{1}{\sqrt{3}}\right)}^4-2{\left(-\frac{1}{\sqrt{3}}\right)}^3-2{\left(-\frac{1}{\sqrt{3}}\right)}^2\\ =\frac{2\sqrt{3}-5}{9}\\ y={\left(-\frac{1}{\sqrt{3}}\right)}^3-\left(-\frac{1}{\sqrt{3}}\right)=\frac{2\sqrt{3}}{9}\\ coordinate:\left(\frac{2\sqrt{3}-5}{9},\frac{2\sqrt{3}}{9}\right)\end{array}$

Similarly

  $\begin{array}{l}t=\frac{1}{\sqrt{3}}\\ x={\left(\frac{1}{\sqrt{3}}\right)}^4-2{\left(\frac{1}{\sqrt{3}}\right)}^3-2{\left(\frac{1}{\sqrt{3}}\right)}^2\\ =-\frac{2\sqrt{3}-5}{9}\\ y={\left(\frac{1}{\sqrt{3}}\right)}^3-\left(\frac{1}{\sqrt{3}}\right)=-\frac{2\sqrt{3}}{9}\\ coordinate:\left(-\frac{2\sqrt{3}-5}{9},-\frac{2\sqrt{3}}{9}\right)\end{array}$

Hence

Vertical tangents are

  $\left(\frac{-3}{16},\frac{3}{8}\right),(0,0),(-8,6)$

Horizontal tangents coordinates are $\left(-\frac{2\sqrt{3}-5}{9},-\frac{2\sqrt{3}}{9}\right),\left(\frac{2\sqrt{3}-5}{9},\frac{2\sqrt{3}}{9}\right)$