Given the function f(x)=x^4−6x^2
a.) Find all values of x where local extrema points occur
b.)find all values of x where inflection points occur
c.) find all local extrema of f(x)
d.) find all global extrema of f(x) over the closed interval [-2,3]
To find the values of x where local extrema points occur, we need to find the critical points of the function. Critical points occur when the derivative of the function is equal to zero or is undefined. Let's start by finding the derivative of f(x).
a.) Finding the derivative of f(x):
f'(x) = 4x^3 - 12x
Next, we need to set the derivative equal to zero and solve for x to find the critical points:
4x^3 - 12x = 0
Factor out 4x:
4x(x^2 - 3) = 0
Set each factor equal to zero and solve for x:
4x = 0 => x = 0
x^2 - 3 = 0 => x^2 = 3 => x = ±√3
So, the critical points are x = 0, x = √3, and x = -√3.
b.) To find the values of x where inflection points occur, we need to find where the second derivative changes sign. Let's find the second derivative of f(x).
Finding the second derivative:
f''(x) = 12x^2 - 12
To determine where the second derivative changes sign, we need to solve the inequality:
12x^2 - 12 > 0
Divide by 12:
x^2 - 1 > 0
Factor the equation:
(x - 1)(x + 1) > 0
The inequality holds true when either both factors are positive or both factors are negative. Considering these two cases:
(i) x - 1 > 0 and x + 1 > 0
x > 1 and x > -1
In this case, x > 1.
(ii) x - 1 < 0 and x + 1 < 0
x < 1 and x < -1
In this case, x < -1.
So, the values of x where inflection points occur are x < -1 and x > 1.
c.) To find the local extrema of f(x), we need to evaluate f(x) at the critical points. Let's substitute the critical points into f(x).
f(0) = 0^4 - 6(0^2) = 0
f(√3) = (√3)^4 - 6(√3)^2 = 3 - 6(3) = 3 - 18 = -15
f(-√3) = (-√3)^4 - 6(-√3)^2 = 3 - 6(3) = 3 - 18 = -15
So, the local extrema points occur at (0, 0), (√3, -15), and (-√3, -15).
d.) To find the global extrema of f(x) over the closed interval [-2,3], we need to evaluate f(x) at the endpoints of the interval and at the critical points obtained in part (c).
f(-2) = (-2)^4 - 6(-2)^2 = 16 - 24 = -8
f(3) = (3)^4 - 6(3)^2 = 81 - 54 = 27
Comparing the values obtained, the global minimum is -15 at points (√3, -15) and (-√3, -15), and the global maximum is 27 at x = 3.