For what values of x does 4x^6-8x^3+18 have a relative minimum
f(x)=4x^6-8x^3+18
The polynomial has a leading coefficient of 4, therefore there is at least one minimum.
A 6th degree polynomial has a maximum of 5 extrema, and a maximum of 3 minima.
The conditions for an extremum at x0 is f'(x0)=0 and f"(x0)≠0.
The conditions for minima are:
f'(x0)=0 and f"(x0)>0.
f(x)=4x^6-8x^3+18
f'(x)=24x^5-24x^2=24x^2(x^3-1)=0
=> f'(x)=0 at x=0 & x=1 (x∈R)
f"(x)=120x^4-48x
f"(0)=0 and f"(1)=120-48=72>0
Therefore f(x) has a single minimum at x=1.
The graph looks like this:
http://img13.imageshack.us/img13/9940/1333888613.png
To determine the values of x for which the function 4x^6 - 8x^3 + 18 has a relative minimum, we need to find the critical points of the function. A relative minimum occurs at a critical point when the derivative changes sign from negative to positive.
1. Start by finding the derivative of the function:
f'(x) = 24x^5 - 24x^2
2. Set the derivative equal to zero and solve for x:
24x^5 - 24x^2 = 0
Factoring out 24x^2, we get:
24x^2(x^3 - 1) = 0
3. Solve for the values of x:
From the first factor, we have x = 0 or x^3 - 1 = 0.
For x^3 - 1 = 0, we can add 1 to both sides and take the cube root to find x:
x = ∛1 = 1
So the critical points are x = 0 and x = 1.
4. To determine the nature of these critical points, we evaluate the second derivative.
f''(x) = 120x^4 - 48x
5. Substitute the values of x into the second derivative.
At x = 0: f''(0) = 120(0)^4 - 48(0) = 0
At x = 1: f''(1) = 120(1)^4 - 48(1) = 72
6. When the second derivative is positive (as in this case f''(1) = 72 > 0), it indicates a relative minimum at that point.
Therefore, the function 4x^6 - 8x^3 + 18 has a relative minimum at x = 1.
In summary, the function has a relative minimum at x = 1.
To find the values of x for which the function has a relative minimum, we need to find the critical points of the function and determine whether each critical point is a relative minimum or maximum.
Here's how you can find the values of x that correspond to the critical points:
1. Take the derivative of the given function:
f'(x) = 24x^5 - 24x^2
2. Set the derivative equal to zero and solve for x to find the critical points:
24x^5 - 24x^2 = 0
Factor out 24x^2:
24x^2(x^3 - 1) = 0
Apply the Zero Product Property:
24x^2 = 0 or x^3 - 1 = 0
Solve each equation separately:
For 24x^2 = 0:
x^2 = 0
x = 0 (double root)
For x^3 - 1 = 0:
x^3 = 1
x = 1 (triple root)
So, we have two critical points: x = 0 and x = 1.
Next, we need to determine whether each critical point represents a relative minimum or maximum. To do this, we can use the second derivative test:
1. Take the second derivative of the function:
f''(x) = 120x^4 - 48x
2. Plug each critical point into the second derivative:
For x = 0:
f''(0) = 120(0)^4 - 48(0) = 0
Since the second derivative at x = 0 is equal to 0, the second derivative test is inconclusive. We cannot determine if x = 0 represents a relative minimum or not.
For x = 1:
f''(1) = 120(1)^4 - 48(1) = 72
Since the second derivative at x = 1 is positive (72 > 0), x = 1 represents a relative minimum.
Therefore, the function 4x^6 - 8x^3 + 18 has a relative minimum at x = 1.