Question

let f(x)=2x^3 - 3x^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 36. 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)

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=2x^{3}-3x^{2}-36x + 2$, then $f'(x)=6x^{2}-6x-36$.

Step2: Solve for horizontal tangent (a)

A horizontal tangent has a slope of 0. Set $f'(x)=0$.
$$6x^{2}-6x - 36=0$$
Divide through by 6: $x^{2}-x - 6=0$.
Factor the quadratic equation: $(x - 3)(x+2)=0$.
Set each factor equal to 0: $x-3 = 0$ gives $x = 3$ and $x + 2=0$ gives $x=-2$.
Find the y - values by substituting x into f(x).
When $x = 3$, $f(3)=2(3)^{3}-3(3)^{2}-36(3)+2=2\times27-3\times9-108 + 2=54-27-108 + 2=-79$.
When $x=-2$, $f(-2)=2(-2)^{3}-3(-2)^{2}-36(-2)+2=2\times(-8)-3\times4 + 72+2=-16-12 + 72+2=46$.
The points are $(3,-79),(-2,46)$.

Step3: Solve for slope of 36 (b)

Set $f'(x)=36$.
$$6x^{2}-6x-36=36$$
Subtract 36 from both sides: $6x^{2}-6x - 72=0$.
Divide through by 6: $x^{2}-x - 12=0$.
Factor the quadratic equation: $(x - 4)(x + 3)=0$.
Set each factor equal to 0: $x-4=0$ gives $x = 4$ and $x + 3=0$ gives $x=-3$.
Find the y - values by substituting x into f(x).
When $x = 4$, $f(4)=2(4)^{3}-3(4)^{2}-36(4)+2=2\times64-3\times16-144 + 2=128-48-144 + 2=-62$.
When $x=-3$, $f(-3)=2(-3)^{3}-3(-3)^{2}-36(-3)+2=2\times(-27)-3\times9 + 108+2=-54-27 + 108+2=29$.
The points are $(4,-62),(-3,29)$.

Answer:

a. $(3,-79),(-2,46)$
b. $(4,-62),(-3,29)$