How do you find the local max and min for # f(x)=x^3-2x+5# on the interval (-2,2)?

1 Answer
Jan 3, 2018

See below.

Explanation:

First find the 1st derivative of #x^3-2x+5#

#d/dx(x^3-2x+5)=3x^2-2#

Maximum and minimum points have a horizontal gradient .i.e. gradient of #0#. Solving the 1st derivative will allow us ti identify these points.

#f'(x)=0#

#:.#

#3x^2-2=0=>x=sqrt(6)/3 , x=-sqrt(6)/3#

Plugging these values into the second derivative, will allow us to find whether these are maximum, minimum or points of inflection. Using the following, if:

#f''(x)>0color(white)(8888)# minimum value

#f''(x)<0color(white)(8888)# maximum value

#f''(x)=0color(white)(8888)# maximum/minimum or point of inflection.

The second derivative is the derivative of the first derivative.

#:.#

#d/dx(3x^2-2)=6x#

Plugging in our values of #x#:

#6(sqrt(6)/3)=2sqrt(6)color(white)(8888888)# ( minimum value )

#6(-sqrt(6)/3)=-2sqrt(6)color(white)(88)# ( maximum value )

Both #color(white)(8888)x=sqrt(6)/3 , x=-sqrt(6)/3# are in #( -2 ,2)#

GRAPH:

graph{y=x^3-2x+5 [-5, 5, -20, 20]}