Math Problem Statement
Solution
This is a calculus problem involving the function . Let us solve it step by step.
Problem Breakdown:
- Given function: .
a) Determine intervals where is increasing or decreasing
To find intervals of increase and decrease, calculate the first derivative and analyze its sign:
Factorize:
Critical points (where ):
Analyze the sign of in the intervals:
- : Choose , → Increasing.
- : Choose , → Decreasing.
- : Choose , → Increasing.
Thus, is:
- Increasing on .
- Decreasing on .
b) Find maximum, minimum points, and inflection points
-
Critical Points: Substitute and into the original function to find the corresponding -values:
- At : . This is a local maximum.
- At : . This is a local minimum.
-
Inflection Points: To find points of inflection, calculate the second derivative:
Solve :
At , test the concavity:
- For (e.g., ): → Concave down.
- For (e.g., ): → Concave up.
Therefore, is an inflection point. Substituting into : Inflection point: .
c) Sketch the graph
The graph should include:
- Critical points: (max), (min).
- Inflection point: .
- Intervals of increase/decrease and concavity.
I'll plot this if you'd like.
Would you like me to plot the graph or explain further?
Related Questions:
- How do you analyze concavity using the second derivative?
- Why do critical points help in identifying maxima and minima?
- What is the significance of inflection points in curve behavior?
- Can you find the x-intercepts of ?
- How does increasing/decreasing behavior relate to derivatives?
Tip:
Always double-check the sign changes of the derivative to avoid mistakes when identifying increasing or decreasing intervals.
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Math Problem Analysis
Mathematical Concepts
Calculus
Critical Points
Intervals of Increase and Decrease
Inflection Points
Formulas
First derivative: y' = 3x^2 - 6x
Second derivative: y'' = 6x - 6
Theorems
Derivative Test for Increasing/Decreasing
Second Derivative Test for Concavity and Inflection Points
Suitable Grade Level
Grades 11-12