Local extrema of the function $f(x)=x^3−3αx^2+3(α^2−1)x+1$

Question

The range of $\alpha$ for which all the points of local extrema of the function $f\left( x \right) = {x^3} - 3\alpha {x^2} + 3\left( {{\alpha ^2} - 1} \right)x + 1$ lie in the interval (–2, 4), is

(A) (–1, 3)

(B) (3, 4)

(C) (–4, –2)

(D) (–2, –1)

My approach is as follow

$f\left( x \right) = {x^3} - 3\alpha {x^2} + 3\left( {{\alpha ^2} - 1} \right)x + 1$

$f'\left( x \right) = 3{x^2} - 6\alpha x + 3\left( {{\alpha ^2} - 1} \right) = 0$

$D = 36{\alpha ^2} - 36\left( {{\alpha ^2} - 1} \right) \geqslant 0$

I can easily solve this problem if I can get the roots of x and I can get the value at which x is at the extremum position, but as I am not getting the value I am struck in this problem.

Answer 1

Note that

\begin{align} f'( x ) & = 3{x^2} - 6\alpha x + 3\left( {{\alpha ^2} - 1} \right)\\ &= 3[x^2 -2ax + (a+1)(a-1)] \\ & = 3[x-(a+1)][x-(a-1)]=0 \end{align}

which leads to the roots $x=a\pm 1$. Then, solve $-2<a\pm 1< 4$ to obtain $a\in (-1,3)$.

Answer 2

Solve the quadratic equation with the parameter $\alpha$:

$x_{1,2}=\frac{6\alpha\pm\sqrt{({36\alpha^2-36\alpha^2+36})}}{6}$

Meaning $x_1=\alpha+1, x_2=\alpha-1$

Can you finish solving from here?