Let f(x)= x^4-4x^3+10
Where are the relative extrema, if they exist? Show all partsof the analysis necessary.
f ' (x) = 4x^3 - 12x^2
= 0 for max/min
4x^2(x-3) = 0
x = 0 or x = 3
take it from there
Yes , I did that part, what else do I have to do?
To find the relative extrema of the function f(x) = x^4 - 4x^3 + 10, we need to start by finding the critical points. Critical points occur where the derivative of the function is either zero or undefined.
1. Find the derivative of f(x):
f'(x) = 4x^3 - 12x^2
2. Set f'(x) equal to zero and solve for x:
4x^3 - 12x^2 = 0
Factor out a common factor of 4x^2:
4x^2(x - 3) = 0
Set each factor equal to zero:
4x^2 = 0 --> x = 0
x - 3 = 0 --> x = 3
So, the critical points are x = 0 and x = 3.
3. Determine the nature of each critical point using the second derivative test. To do this, we need to find the second derivative of f(x).
Taking the derivative of f'(x), we get:
f''(x) = 12x^2 - 24x
4. Now, substitute the critical points (x = 0 and x = 3) into the second derivative to determine the nature of each critical point.
When x = 0:
f''(0) = 12(0)^2 - 24(0) = 0
When x = 3:
f''(3) = 12(3)^2 - 24(3) = 72
5. Analyzing the results:
- When f''(x) = 0, the second derivative test is inconclusive.
- When f''(x) > 0, the point is a relative minimum.
- When f''(x) < 0, the point is a relative maximum.
Based on our analysis:
- At x = 0, the nature of the critical point is inconclusive.
- At x = 3, the nature of the critical point is a relative minimum.
Therefore, the function has a relative minimum at x = 3.