Given the function:
[tex]h(x)=x^3-4x^2-7x+12[/tex]
Find the first derivative:
[tex]h^{\prime}(x)=3x^2-8x-7[/tex]
The first derivative gives us the slope of the tangent line to the graph of the function. When the tangent line is horizontal, the slope is 0, thus:
[tex]3x^2-8x-7=0[/tex]
This is a quadratic equation with coefficients a = 3, b = -8, c = -7.
To calculate the solutions to the equation, we use the quadratic solver formula:
[tex]$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$
[/tex]
Substituting:
[tex]x=\frac{-(-8)\pm\sqrt{(-8)^2-4(3)(-7)}}{2(3)}[/tex]
Operate:
[tex]\begin{gathered} x=\frac{8\pm\sqrt{64+84}}{6} \\ \\ x=\frac{8\pm\sqrt{148}}{6} \end{gathered}[/tex]
Since:
[tex]148=2^2\cdot37[/tex]
We have:
[tex]\begin{gathered} x=\frac{8\pm2\sqrt{37}}{6} \\ \\ \text{ Simplifying by 2:} \\ \\ x=\frac{4\pm\sqrt{37}}{3} \end{gathered}[/tex]
There are two solutions:
[tex]\begin{gathered} x_1=\frac{4+\sqrt{37}}{3} \\ \\ x_2=\frac{4-\sqrt{37}}{3} \end{gathered}[/tex]
