find the critical points, inflection points, the absolute minimum value of y, and the relative maximum points of y=x^4-3x^2+2
dy/dx = 4x^3 - 6x
= 0 for max/mins
2x(2x^2 - 3) = 0
x = 0 or x = ±√(3/2)
if x = 0, y = 2
if x = +√(3/2) y = ....
if x = -√(3/2) y = .... ----- you do the arithmetic
y'' = 12x^2 - 6 = 0 for points of inflection
x^2 = 6/12 = 1/2
x = ±√(1/2)
sub back in to get the two points of inflection
I ran it through Wolfram to see the shape of the graph
http://www.wolframalpha.com/input/?i=y%3Dx%5E4-3x%5E2%2B2
To find the critical points, inflection points, absolute minimum value, and relative maximum points of the function y = x^4 - 3x^2 + 2, follow these steps:
Step 1: Find the derivative of the function.
To find the critical points, we need to find where the derivative of the function is equal to zero or undefined. Take the derivative of y with respect to x:
y' = 4x^3 - 6x
Step 2: Find the critical points.
Set the derivative equal to zero and solve for x:
4x^3 - 6x = 0
Factor out x:
x(4x^2 - 6) = 0
Set each factor equal to zero:
x = 0 or 4x^2 - 6 = 0
For x = 0, we have a critical point.
Solve the second equation:
4x^2 - 6 = 0
4x^2 = 6
x^2 = 6/4
x^2 = 3/2
Taking the square root of both sides, we get:
x = ±√(3/2)
So we have two more critical points: x = √(3/2) and x = -√(3/2).
Step 3: Find the second derivative.
To determine the inflection points, we need to find where the second derivative changes its sign. Take the derivative of y' with respect to x:
y'' = 12x^2 - 6
Step 4: Find the inflection points.
Set the second derivative equal to zero and solve for x:
12x^2 - 6 = 0
12x^2 = 6
x^2 = 6/12
x^2 = 1/2
Taking the square root of both sides, we get:
x = ±√(1/2)
So we have two possible inflection points: x = √(1/2) and x = -√(1/2).
Step 5: Evaluate the function at critical points and determine the minimum and maximum.
Now, let's plug these critical points and inflection points into the original function y = x^4 - 3x^2 + 2 and find the corresponding y-values:
At x = 0: y = (0)^4 - 3(0)^2 + 2 = 0 - 0 + 2 = 2
At x = √(3/2): y = (√(3/2))^4 - 3(√(3/2))^2 + 2
At x = -√(3/2): y = (-√(3/2))^4 - 3(-√(3/2))^2 + 2
At x = √(1/2): y = (√(1/2))^4 - 3(√(1/2))^2 + 2
At x = -√(1/2): y = (-√(1/2))^4 - 3(-√(1/2))^2 + 2
Step 6: Analyze the results.
The critical points are x = 0, x = √(3/2), and x = -√(3/2).
The inflection points are x = √(1/2) and x = -√(1/2).
The absolute minimum value of y is 2.
To determine the relative maximum points, we need to compare the y-values at the critical points. In this case, y = 2 is the only critical point that evaluates to a maximum value.
Therefore, the critical points are x = 0, x = √(3/2), and x = -√(3/2). The inflection points are x = √(1/2) and x = -√(1/2). The absolute minimum value of y is 2, and the relative maximum point is (0, 2).