Answer:
(1, 1/2) and (7, 1/14)
Explanation:
Given:
[tex]f(x)=\frac{x-4}{x^2-7}[/tex]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.
[tex]f^{\prime}(x)=\frac{vu^{\prime}-uv^{\prime}}{v^2}[/tex]From f(x):
[tex]\begin{gathered} u=x-4\implies\frac{du}{dx}=1 \\ v=x^2-7\implies\frac{dv}{dx}=2x \end{gathered}[/tex]Substitute into the quotient rule:
[tex]\begin{gathered} f^{\prime}(x)=\frac{(x^2-7)(1)-2x(x-4)}{(x^2-7)^2} \\ =\frac{x^2-7-2x^2+8x}{(x^2-7)^2} \\ f^{\prime}(x)=\frac{-x^2+8x-7}{(x^2-7)^2} \end{gathered}[/tex]Set the derivative equal to 0:
[tex]\begin{gathered} \frac{-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}[/tex]Finally, find the corresponding values of f(x).
[tex]\begin{gathered} \text{ At x=1: }f(1)=\frac{1-4}{1^2-7}=\frac{-3}{-6}=0.5 \\ \text{At x=7: }f(7)=\frac{7-4}{7^2-7}=\frac{3}{42}=\frac{1}{14} \end{gathered}[/tex]The points at which the graph of the function has a horizontal tangent line are:
[tex]\begin{gathered} (1,\frac{1}{2}) \\ (7,\frac{1}{14}) \end{gathered}[/tex]