Respuesta :
Solution:
Given the function f(x) expressed as
[tex]f(x)=x^3-4x^2-2x+7[/tex]For horizontal tangent, we have
[tex]f^{\prime}(x)=0[/tex]Thus, we have
[tex]\begin{gathered} f^{\prime}(x)=3x^2-8x-2 \\ when\text{ f'\lparen x\rparen=0, we have} \\ 3x^2-8x-2=0 \\ \end{gathered}[/tex]solving for x using quadratic formula, we have
[tex]\begin{gathered} x=\frac{-\left(-8\right)\pm\:2\sqrt{22}}{2\cdot\:3} \\ \Rightarrow x=\frac{-\left(-8\right)+2\sqrt{22}}{2\cdot\:3} \\ =\frac{4+\sqrt{22}}{3} \\ or \\ \Rightarrow x=\frac{-(-8)2\sqrt{22}}{2\times3} \\ =\frac{4-\sqrt{22}}{3} \end{gathered}[/tex]Hence, the solution are values of x that satisfy f'(x) = 0. That is, solve the equation
[tex]3x^2-8x-2=0[/tex]x=
[tex]\frac{4+\sqrt{22}}{3},\frac{4-\sqrt{22}}{3}[/tex]