And Justify True OR False f(a, b) = (a-2)² + (6-3)³. Then, we say Pornt (2.31 T f=R²-)R. is local min of f over R²
And Justify True OR False f(a, b) = (a-2)² + (6-3)³. Then, we say Pornt (2.31 T f=R²-)R. is local min of f over R²
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![### True or False and Justify
Consider the function \( f(a, b) = (a - 2)^2 + (b - 3)^3 \), where \( f: \mathbb{R}^2 \rightarrow \mathbb{R} \).
We need to determine if the statement "The point \((2, 3)^T\) is a local minimum of the function \( f \) over \(\mathbb{R}^2\)" is true or false.
#### Analysis
The given function is:
\[ f(a, b) = (a - 2)^2 + (b - 3)^3 \]
To analyze if the point \((2, 3)^T\) is a local minimum, let's consider the nature of the function and the behavior of each term:
1. The term \((a - 2)^2\) is always non-negative and equals zero when \(a = 2\).
2. The term \((b - 3)^3\) changes sign as \(b\) crosses 3 and has a critical point at \(b = 3\), where its value is also zero.
In order to have a local minimum at \((2, 3)\), both the first and second partial derivatives of \(f\) need to satisfy certain conditions. Specifically, the gradient \( \nabla f \) should be zero at \((2, 3)\) and the Hessian matrix should be positive definite.
#### Gradient Calculation
Calculate the partial derivatives:
\[
\frac{\partial f}{\partial a} = 2(a - 2)
\]
\[
\frac{\partial f}{\partial b} = 3(b - 3)^2
\]
Evaluating at \((a, b) = (2, 3)\):
\[
\frac{\partial f}{\partial a}\bigg|_{(2, 3)} = 2(2 - 2) = 0
\]
\[
\frac{\partial f}{\partial b}\bigg|_{(2, 3)} = 3(3 - 3)^2 = 0
\]
Therefore, \( \nabla f(2, 3) = (0, 0) \), so the gradient condition is satisfied.
#### Hessian Matrix Calculation
Calculate the second partial derivatives to form](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e71c6f0-fdac-4001-9095-2485c121cf1d%2F4d082fea-3ff0-49af-8dc5-84290b5f877f%2F40iobld_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### True or False and Justify
Consider the function \( f(a, b) = (a - 2)^2 + (b - 3)^3 \), where \( f: \mathbb{R}^2 \rightarrow \mathbb{R} \).
We need to determine if the statement "The point \((2, 3)^T\) is a local minimum of the function \( f \) over \(\mathbb{R}^2\)" is true or false.
#### Analysis
The given function is:
\[ f(a, b) = (a - 2)^2 + (b - 3)^3 \]
To analyze if the point \((2, 3)^T\) is a local minimum, let's consider the nature of the function and the behavior of each term:
1. The term \((a - 2)^2\) is always non-negative and equals zero when \(a = 2\).
2. The term \((b - 3)^3\) changes sign as \(b\) crosses 3 and has a critical point at \(b = 3\), where its value is also zero.
In order to have a local minimum at \((2, 3)\), both the first and second partial derivatives of \(f\) need to satisfy certain conditions. Specifically, the gradient \( \nabla f \) should be zero at \((2, 3)\) and the Hessian matrix should be positive definite.
#### Gradient Calculation
Calculate the partial derivatives:
\[
\frac{\partial f}{\partial a} = 2(a - 2)
\]
\[
\frac{\partial f}{\partial b} = 3(b - 3)^2
\]
Evaluating at \((a, b) = (2, 3)\):
\[
\frac{\partial f}{\partial a}\bigg|_{(2, 3)} = 2(2 - 2) = 0
\]
\[
\frac{\partial f}{\partial b}\bigg|_{(2, 3)} = 3(3 - 3)^2 = 0
\]
Therefore, \( \nabla f(2, 3) = (0, 0) \), so the gradient condition is satisfied.
#### Hessian Matrix Calculation
Calculate the second partial derivatives to form
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