What is the relative minimum, relative maximum, and points of inflection of #f(x) = x^4 - 4x^2#?

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Answer 1 · 2016-11-24T20:41:54

#f'(x)=4x^3-8x#
#f'(x)=4x(x^2-2)#

#4x(x^2-2)=0#

#x=0, x=+-sqrt2#

The multiplicity of each of these zeros is odd, therefore, there will be a minimum or maximum at each value.

For:
#x>sqrt2#
#y>0#
Positive slope

For:
#0 < x < sqrt2#
#y < 0#
Negative slope

For:
#-sqrt2 < x < 0#
#y > 0#
Positive slope

For :
#x < -sqrt2#
#y < 0#
Negative slope

#:. x=sqrt2 =># Local minimum
#:. x=0 =># Local maximum
#:. x=-sqrt2 =># Local minimum

#f''(x)=12x^2-8#
#f''(x)=4(3x^2-2)#

#3x^2-2=0#
#x=+-sqrt(2/3)#

For:
#x>sqrt(2/3)#
#y>0#
Concave up

For:
#-sqrt(2/3) < x < sqrt(2/3)#
#y<0#
Concave down

For:
#x < -sqrt(2/3)#
#y>0#
Concave up

#:.# There are points of inflection at:
#x=+-sqrt(2/3)#