To find the points at which the tangent line is horizontal, it is enough to find where the slope of the function is 0 because a horizontal line's slope is 0.
Find f'(x).
[tex]f^{\prime}(x)=3x^2-9[/tex]
Now set it equal to 0 and solve for x to find the x values at which the tangent line is horizontal to given function.
[tex]\begin{gathered} 3x^2-9=0 \\ 3x^2=9 \\ x^2=3 \\ x=\pm\sqrt[]{3} \end{gathered}[/tex]
Find the function values at these x-values.
[tex]\begin{gathered} f(\sqrt[]{3})=(\sqrt[]{3})^3-9\sqrt[]{3} \\ =3\sqrt[]{3}-9\sqrt[]{3} \\ =3\sqrt[]{3}(1-3) \\ =-6\sqrt[]{3} \end{gathered}[/tex]
[tex]\begin{gathered} f(-\sqrt[]{3})=(-\sqrt[]{3})^3+9\sqrt[]{3} \\ =-3\sqrt[]{3}+9\sqrt[]{3} \\ =3\sqrt[]{3}(-1+3) \\ =6\sqrt[]{3} \end{gathered}[/tex]
The points where the graph of the function have horizontal tangents are
[tex](\sqrt[]{3},-6\sqrt[]{3}),(-\sqrt[]{3},6\sqrt[]{3})[/tex]
Option A is correct.
