Question

Horizontal Tangent Line, determine the point(s) at which the graph of the function has a horizontal tangent line.

Answer 1

(1, 1/2) and (7, 1/14)

Explanation:

Given:

`f(x)=(x-4)/(x^2-7)`

The point(s) at which the graph of the function has a horizontal tangent line is the point at which the first derivative is zero.

First, find the first derivative of f(x) using the quotient rule.

`f^(\prime)(x)=(vu^(\prime)-uv^(\prime))/(v^2)`

From f(x):

`\begin{gathered} u=x-4\implies(du)/(dx)=1 \\ v=x^2-7\implies(dv)/(dx)=2x \end{gathered}`

Substitute into the quotient rule:

`\begin{gathered} f^(\prime)(x)=((x^2-7)(1)-2x(x-4))/((x^2-7)^2) \\ =(x^2-7-2x^2+8x)/((x^2-7)^2) \\ f^(\prime)(x)=(-x^2+8x-7)/((x^2-7)^2) \end{gathered}`

Set the derivative equal to 0:

`\begin{gathered} (-x^2+8x-7)/(\left(x^2-7\right))=0\implies-x^2+8x-7=0 \\ \text{ Solve for x:} \\ -x^2+7x+x-7=0 \\ -x(x-7)+1(x-7)=0 \\ (x-7)(-x+1)=0 \\ x-7=0\text{ or }-x+1=0 \\ x=7,x=1 \end{gathered}`

Finally, find the corresponding values of f(x).

`\begin{gathered} \text{ At x=1: }f(1)=(1-4)/(1^2-7)=(-3)/(-6)=0.5 \\ \text{At x=7: }f(7)=(7-4)/(7^2-7)=(3)/(42)=(1)/(14) \end{gathered}`

The points at which the graph of the function has a horizontal tangent line are:

`\begin{gathered} (1,(1)/(2)) \\ (7,(1)/(14)) \end{gathered}`