## Calculus with Applications (10th Edition)

$$x = \frac{{4 + \sqrt {37} }}{3}{\text{ and }}x = \frac{{4 - \sqrt {37} }}{3}$$
\eqalign{ & f\left( x \right) = {x^3} - 4{x^2} - 7x + 8 \cr & {\text{find the derivative of the function}} \cr & f'\left( x \right) = {D_x}\left( {{x^3} - 4{x^2} - 7x + 8} \right) \cr & {\text{use the power rule for derivatives}} \cr & f'\left( x \right) = 3{x^2} - 4\left( {2x} \right) - 7\left( 1 \right) \cr & f'\left( x \right) = 3{x^2} - 8x - 7 \cr & \cr & {\text{Find the points of }}x{\text{ where the slope of the tangent line is horizontal}}{\text{, set }}f'\left( x \right) = 0 \cr & {\text{Then}} \cr & 3{x^2} - 8x - 7 = 0 \cr & {\text{use the quadratic formula }}x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \cr & x = \frac{{ - \left( { - 8} \right) \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( 3 \right)\left( { - 7} \right)} }}{{2\left( 3 \right)}} \cr & {\text{simplifying}} \cr & x = \frac{{8 \pm \sqrt {148} }}{6} \cr & x = \frac{{8 \pm 2\sqrt {37} }}{6} \cr & x = \frac{{4 \pm \sqrt {37} }}{3} \cr & {\text{then}} \cr & x = \frac{{4 + \sqrt {37} }}{3}{\text{ and }}x = \frac{{4 - \sqrt {37} }}{3} \cr}