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
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
To find the critical points of the function, we need to find the values of x where the derivative is equal to zero or is undefined.
The derivative of the given function is -8x^3 + 6x. We set this equal to zero and solve for x:
-8x^3 + 6x = 0
Let's factor out an x:
x(-8x^2 + 6) = 0
Set each factor equal to zero:
x = 0
-8x^2 + 6 = 0
To solve -8x^2 + 6 = 0, we isolate x:
-8x^2 = -6
Divide both sides by -8:
x^2 = 3/4
Take the square root of both sides:
x = ±√(3/4)
Simplifying further:
x = ±√(3/4) = ±√3/2
Therefore, the critical points are x = 0, x = √3/2, and x = -√3/2.
To determine which critical points are extrema, we can use the Second Derivative Test. However, since we only have one critical point that is not at a boundary (x = 0), we can just evaluate the function at that point and check if it is a local minimum or maximum:
For x = 0:
y = -2(0)^4 + 3(0)^2
y = 0
Since the y-value at x = 0 is 0, we can conclude that it is a local minimum.
Therefore, the critical points are x = 0 (local minimum), x = √3/2, and x = -√3/2.