Minimization
In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in terms of global or local range. The definition of minimization in the thesaurus is the process of reducing something to a small amount, value, or position. Minimization (noun) is an instance of belittling or disparagement.
Maxima and Minima
The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.
Derivatives
A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which one would be easier? Walking on the road would be much easier than climbing a mountain.
Concavity
In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the first derivative test and second derivative test to understand the concave behavior of the function.
![**Problem Statement:**
Given the function \( f(x) = 2x^3 + 6x^2 \), determine all intervals on which \( f \) is both decreasing and concave up.
**Explanation:**
To solve this problem, follow these steps:
1. **Find the first derivative \( f'(x) \)**: This will help identify where the function is increasing or decreasing.
\[
f'(x) = \frac{d}{dx}(2x^3 + 6x^2) = 6x^2 + 12x
\]
2. **Find the critical points** by setting \( f'(x) = 0 \):
\[
6x^2 + 12x = 0 \implies 6x(x + 2) = 0
\]
The solutions are \( x = 0 \) and \( x = -2 \).
3. **Find the second derivative \( f''(x) \)**: This will help determine where the function is concave up or down.
\[
f''(x) = \frac{d}{dx}(6x^2 + 12x) = 12x + 12
\]
4. **Determine intervals** where \( f''(x) > 0 \) (concave up):
\[
12x + 12 > 0 \implies x > -1
\]
5. **Combine information**:
- The function is **decreasing** where \( f'(x) < 0 \), which is between the critical points: \( x \in (-\infty, -2) \).
- The function is **concave up** where \( x > -1 \).
6. **Find common intervals**:
- The intersection of \( x \in (-\infty, -2) \) (decreasing interval) and \( x > -1 \) (concave up interval) is none since there are no overlapping regions.
Therefore, there are no intervals where \( f(x) = 2x^3 + 6x^2 \) is both decreasing and concave up.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc88a8d3c-414f-44f0-8ef5-044fbd549dcd%2F132b162d-65f5-45b9-8e5d-8ad3a475c91b%2Fin9sp1_processed.jpeg&w=3840&q=75)
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