Question

let ( f(x)=2x^{3}-x^{2}-36x + 2). a. find all points on the graph of ( f(x)) at which the tangent line is horizontal. b. find all points on the graph of ( f(x)) at which the tangent line has slope 30. a. the tangent line is horizontal at the point(s) . (type an ordered - pair. use a comma to separate answers as needed.)

Explanation:

Step1: Find the derivative of $f(x)$

The power - rule for differentiation states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. Given $f(x)=2x^{3}-x^{2}-36x + 2$, then $f^\prime(x)=6x^{2}-2x-36$.

Step2: For horizontal tangent line

A horizontal tangent line has a slope of 0. So we set $f^\prime(x)=0$.
$$6x^{2}-2x - 36=0$$
Divide through by 2: $3x^{2}-x - 18=0$.
Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 3$, $b=-1$, and $c=-18$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-1)^{2}-4\times3\times(-18)=1 + 216=217$.
Then $x=\frac{1\pm\sqrt{217}}{6}$.
When $x=\frac{1+\sqrt{217}}{6}$, $y = 2(\frac{1+\sqrt{217}}{6})^{3}-(\frac{1+\sqrt{217}}{6})^{2}-36(\frac{1+\sqrt{217}}{6})+2$.
When $x=\frac{1 - \sqrt{217}}{6}$, $y = 2(\frac{1-\sqrt{217}}{6})^{3}-(\frac{1-\sqrt{217}}{6})^{2}-36(\frac{1-\sqrt{217}}{6})+2$.

Answer:

$(\frac{1+\sqrt{217}}{6},2(\frac{1+\sqrt{217}}{6})^{3}-(\frac{1+\sqrt{217}}{6})^{2}-36(\frac{1+\sqrt{217}}{6})+2),(\frac{1 - \sqrt{217}}{6},2(\frac{1-\sqrt{217}}{6})^{3}-(\frac{1-\sqrt{217}}{6})^{2}-36(\frac{1-\sqrt{217}}{6})+2)$