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 )
given; f(x)=x^3 -4x ^2 - 11x+30
calculate the coodinates of the stationary points of f
1 answer