Prove that the function $f(x)=x^3+3x^2+x+a$,$a$ being a constant ,is monotonic decreasing in $[\alpha,\beta]$

Question

Prove that the function $f(x)=x^3+3x^2+x+a$,$a$ being a constant ,is monotonic decreasing in $[\alpha,\beta]$ where $x=\alpha$ and $x=\beta$ are two successive points of extremum of the function.

I tried to solve this question.
$f'(x)=3x^2+6x+1=0$ gives $x=\frac{-6\pm2\sqrt{6}}{6}=\frac{-6+2\sqrt{6}}{6},\frac{-6-2\sqrt{6}}{6}$

$f''(x)=6x+6$ is negative for both the critical points.So both are points of maxima.What to do further to prove the question?Please help me,i got stuck.

Answer 1

As Svetoslav points out in the comments you just miscalculated. If $x_1 = \frac{-6+2\sqrt{6}}{6}$ and $x_2 = \frac{-6-2\sqrt{6}}{6}$ then $f''(x_1) = 6 x_1 + 6 = 2 \sqrt{6} > 0$ thus $x_1 \approx -0.184$ is a local minimum and and likewise $f''(x_2) = - 2 \sqrt{6} < 0$ and thus $x_2 \approx -1.816$ is a local maximum.