Question

given; f(x)=x^3 -4x ^2 - 11x+30

calculate the coodinates of the stationary points of f

Answer 1

A stationary point is a point at which first derivative f(x) ' = 0

In this case:

f(x) ' = ( x³ - 4 x² - 11 x + 30 )' = 3 x² - 4 ∙ 2 x - 11

f(x) ' = 3 x² - 8 x - 11

Now you must find where:

f(x) ' = 0

3 x² - 8 x - 11 = 0

The solutions are:

x = - 1

and

x = 11 / 3

Now you must find second derivative f(x)"

f(x)" = [ f(x) ' ] = ( 3 x² - 8 x - 11 ) ' = 3 ∙ 2 x - 8

f(x)" = 6 x - 8

If f''(x) < 0, the stationary point at x is concave down; a maximal extremum

If f''(x) > 0, the stationary point at x is concave up; a minimal extremum

If f''(x) = 0, the nature of the stationary point must be determined by way of other means, often by noting a sign

In this case:

f( - 1 )" = 6 ∙ ( - 1 ) - 8 = - 6 - 8 = - 14 < 0

For x = - 1 , the stationary point at x is concave down; a maximal extremum

f( 11 / 3 )" = 6 ∙ ( 11 / 3 ) - 8 = 22 - 8 = 14 > 0

For x = 11 / 3 , the stationary point at x is concave up; a minimal extremum

Stationary pointa are:

Minimum:

x = 11 / 3 , f(11/3) = ( 11 / 3 ) ³ - 4 ∙ ( 11 / 3 )² - 11 ∙ 11 / 3 + 30 = - 400 / 27

( 11 / 3 , - 400 / 27 )

Maximum:

x = - 1 , f(-1) = ( - 1 )³ - 4 ∙ ( - 1 )² - 11 ∙ ( - 1 ) + 30 = 36

( - 1 , 36 )