Question
Find all critical points and determine which critical points are extrema by testing intervals.
y=-2x^4+3x^2
I found the derivative which is:
-8x^3+6x
y=-2x^4+3x^2
I found the derivative which is:
-8x^3+6x
Answers
Damon
yes, but where is it zero?
function is horizontal when
-8x^3+6x = 0
that happens when
x (-8 x^2 + 6) = 0
x (-4x^2 + 3) = 0
x = 0
or
x^2 = 3/4
x = + or - (1/2)sqrt (3)
--------------------
now what is second derivative at those points?
d^2y/dx^2 = -24 x^2 + 6
when x = 0, that is positive so it is a MINIMUM
when x^2 = 3/4
d^2y/dx^2 = -18 + 6 = -12
so it is a MAXIMUM at those two oints
function is horizontal when
-8x^3+6x = 0
that happens when
x (-8 x^2 + 6) = 0
x (-4x^2 + 3) = 0
x = 0
or
x^2 = 3/4
x = + or - (1/2)sqrt (3)
--------------------
now what is second derivative at those points?
d^2y/dx^2 = -24 x^2 + 6
when x = 0, that is positive so it is a MINIMUM
when x^2 = 3/4
d^2y/dx^2 = -18 + 6 = -12
so it is a MAXIMUM at those two oints