The point (-1, 7) The point (1, -5)

Given that :

The function is f(x) = 3x³ - 9x + 1.

By using,

Suppose x = c is the critical point of f(x) then,

If f ^' (x ) > 0 to the left of x = c and f ^' (x) < 0 to the right of x = c , then x = c is a relative maximum .

If f ^' (x ) < 0 to the left of x = c and f ^' (x) > 0 to the right of x = c , then x = c is a relative minimum .

An inflection point is a point on the graph at which the second derivative changes sign.

If f ^'' ( x ) > 0 then f(x) concave upwards.

If f^'' (x) < 0 then f(x) concave downwards.

To find the critical points :

Differentiate the given equation with respect to x,

f^' (x) = 9 x² - 9

Set f^' (x) = 0

9 x² - 9 = 0

9 ( x² - 1) = 0

x² - 1 = 0

x² = 1

Solve for x:

x = - 1, x = 1.

The critical points are x = - 1 and x = 1.

The domain of the given function is $-\infty < x< \infty$ .

Combine the critical points x = - 1, x = 1 with the domain.

The functions monotone intervals are $- \infty < x< -1 , - 1< x< 1, 1 < x< \infty$ .

To check the sign of  f^' (x) = 9 x² - 9 at each monotone interval :

$\begin{array}{cccccc}& -\infty < x<-1& x = -1& -1<x<1& x=1& 1<x<\infty \\ Sign& +& 0& -& 0& +\\ Behavior& Increa\sin g& Maximum& Dcrea\sin g& Minimum& Increa\sin g\end{array}$

Plug the extreme points x = - 1 in the given function.

f(-1) = 7 , to get y = 7.

Relative maximum = ( -1, 7)