Question

Let p(x)=ax3−x, where a is constant and a>0.


Find the local maxima and minima of p.

(Enter your maxima and minima as comma-separated xvalue,classification pairs. For example, if you found that x=−2 was a local minimum and x=3 was a local maximum, you should enter (-2,min), (3,max). If there were no maximum, you must drop the parentheses and enter -2,min.)


maxima and minima:

Answer 1

Apply the First Sufficient Condition For The Extremum. Find the first derivative.

$p'(x) = 3ax^2-1$

Find x_{0 }where the first derivative is equal to zero.

$3ax^2-1=0$

$x^2=x_1=\frac{1}{3a}$

$x_1=\sqrt{\frac{1}{3a}}$ and $x_2=-\sqrt{\frac{1}{3a}}$

Check that $p'(x)$ changes the sign in x_{1 }and x_{2 .} $p'(x)$ is a parabola with vertex $x_0=\frac{-b} {2a}=0; y_0=p'(x_0)=-1$

It means that parabola branches cross the O_{x }in x_{1 }and x_{2 . }

The graph of derivative is shown on Picture 1:

Picture 1

As we can see the vertex is in 3 and 4 quarters where the function has minus sign and after crossing the O_x in x₁ and x₂ function goes to 1 and 2 quarters and changes the sign to plus.

It means that extrema are found.

Find the value of function p(x) in extrema and compare.

$p(x_1)=ax_1^3-x_1=-\frac{2}{3}*\sqrt{\frac{1}{3a}}$

$p(x_2)=\frac{2}{3}*\sqrt{\frac{1}{3a}}$

As $p(x_2)>p(x_1)$ then x₂ is a point of maximum and x₁ is a point of minimum.

Answer

$(\sqrt{\frac{1}{3a}},min)$ 

$(-\sqrt{\frac{1}{3a}},max)$