Final answer:
In analyzing the function f(x) = x^5 − 2x^3, you calculate the first and second derivatives to find intervals of increase, decrease, concavity and inflection points. The critical points for local extrema are given by the first derivative, while inflection points are found via the second derivative.
Step-by-step explanation:
Analysis of the Function f(x) = x5 − 2x3
For the function f(x) = x5 − 2x3, we need to determine the intervals of increase and decrease, concavity, and find all local extrema and inflection points.
Intervals of Increase and Decrease
To find where the function is increasing or decreasing, we calculate the first derivative f'(x) and analyze its sign:
f'(x) = 5x4 - 6x2
Setting f'(x) to zero to find critical points:
0 = 5x4 - 6x2
This gives us x = 0, x = √(6/5), and x = -√(6/5) as critical points. Testing intervals around these points will determine where f(x) is increasing and decreasing.
Concavity and Inflection Points
To determine concavity, we calculate the second derivative f''(x):
f''(x) = 20x3 - 12x
Setting f''(x) to zero for inflection points:
0 = 20x3 - 12x
We get x = 0 and x = √(3/5), and x = -√(3/5) as potential inflection points. Again, testing intervals around these points will help us determine where the function is concave up or down.
Local Maxima and Minima
Using the first derivative test around the critical points, we identify the location of any local maxima and minima.
Increasing Intervals: Where f'(x) > 0
Decreasing Intervals: Where f'(x) < 0
Concave Up: Where f''(x) > 0
Concave Down: Where f''(x) < 0
Local Maxima/Minima: The location of maxima/minima is where the sign of f'(x) changes from positive to negative/negative to positive, respectively.
Inflection Points: Occur where the concavity changes, at x = √(3/5) and x = -√(3/5).