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Magazine Chapter 3: Problem 8
Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$ g(z)=\frac{z^{2}}{1+z^{2}} $$
Short Answer
Expert verified
The critical point \( z = 0 \) is a local minimum by both tests.
Step by step solution
01
Find the first derivative
To find the critical points, we first need the derivative of the function \( g(z) = \frac{z^2}{1+z^2} \). Use the quotient rule for derivatives: if \( f(z) = \frac{u(z)}{v(z)} \), then \( f'(z) = \frac{u'(z)v(z) - u(z)v'(z)}{[v(z)]^2} \). For \( g(z) \), let \( u(z) = z^2 \) and \( v(z) = 1+z^2 \). This gives \( u'(z) = 2z \) and \( v'(z) = 2z \). Now, apply the quotient rule:\[ g'(z) = \frac{2z(1+z^2) - z^2(2z)}{(1+z^2)^2} = \frac{2z + 2z^3 - 2z^3}{(1+z^2)^2} \ldots ...= \frac{2z}{(1+z^2)^2} \].
02
Find critical points
Critical points occur where the derivative equals zero or is undefined. Set \( g'(z) = 0 \):\[ \frac{2z}{(1+z^2)^2} = 0 \].The numerator determines when the fraction is zero, so \( 2z = 0 \). Solve for \( z \) to find the critical point:\( z = 0 \). Since the denominator is never zero (\( 1+z^2 \geq 1 \)), the derivative is defined for all \( z \). Thus, the only critical point is \( z = 0 \).
03
Apply the First Derivative Test
To apply the First Derivative Test, analyze the sign of \( g'(z) \) around the critical point \( z = 0 \). Choose test values such as \( z = -1 \) and \( z = 1 \):- For \( z = -1 \), \( g'(-1) = \frac{2(-1)}{(1+(-1)^2)^2} = \frac{-2}{4} = -0.5 \). The derivative is negative.- For \( z = 1 \), \( g'(1) = \frac{2(1)}{(1+1^2)^2} = \frac{2}{4} = 0.5 \). The derivative is positive.Since \( g'(z) \) changes from negative to positive as \( z \) increases through 0, by the First Derivative Test, \( z = 0 \) is a local minimum.
04
Find the second derivative
To perform the Second Derivative Test, find \( g''(z) \). Differentiate \( g'(z) = \frac{2z}{(1+z^2)^2} \) using the product and chain rules:Let \( h(z) = (1+z^2)^2 \) and \( h'(z) = 4z(1+z^2) \). The derivative of \( g'(z) \) is:\[ g''(z) = \left( \frac{d}{dz} \right) \frac{2z}{h(z)} = \frac{2h(z) - 2zh'(z)}{h(z)^2} \].Simplifying yields:\[ g''(z) = \frac{2(1+z^2)^2 - 8z^2(1+z^2)}{(1+z^2)^4} = \frac{2(1+z^2) - 8z^2}{(1+z^2)^3} = \frac{2 - 6z^2}{(1+z^2)^3} \].
05
Apply the Second Derivative Test
Evaluate \( g''(z) \) at the critical point \( z = 0 \):\[ g''(0) = \frac{2 - 6(0)^2}{(1+0^2)^3} = 2 \].Since \( g''(0) > 0 \), the function is concave up at \( z = 0 \), indicating a local minimum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative Test
The First Derivative Test helps us determine if a function has a local maximum or minimum at a critical point. A critical point occurs where the first derivative of the function is zero or undefined.
To use this test, follow these steps:
Setting this equal to zero, we found a critical point at \(z = 0\). By selecting test values \(z = -1\) and \(z = 1\), we determined that the first derivative changes from negative to positive as it moves through \(z = 0\). This sign change indicates a local minimum at \(z = 0\).
To use this test, follow these steps:
- Calculate the first derivative of the function.
- Identify and solve for the critical points by setting the derivative equal to zero or checking where it is undefined.
- Choose values slightly less and slightly greater than each critical point. These values will help you assess the derivative's sign change.
- Evaluate the derivative at these test values to see if it shifts from positive to negative or vice versa.
Setting this equal to zero, we found a critical point at \(z = 0\). By selecting test values \(z = -1\) and \(z = 1\), we determined that the first derivative changes from negative to positive as it moves through \(z = 0\). This sign change indicates a local minimum at \(z = 0\).
Second Derivative Test
The Second Derivative Test can sometimes be used to further verify the nature of a critical point by examining the concavity of the function. Here's how it works:
Evaluating this at \(z = 0\) gives \(g''(0) = 2\), which is positive. Therefore, it confirms the presence of a local minimum at \(z = 0\).
- First, ensure you have a critical point from the First Derivative Test.
- Find the second derivative of the original function.
- Plug the critical point into the second derivative to evaluate it.
- If the second derivative is positive, the function is concave up at the critical point, indicating a local minimum.
- If the second derivative is negative, the function is concave down, suggesting a local maximum.
- If the second derivative equals zero, the test is inconclusive.
Evaluating this at \(z = 0\) gives \(g''(0) = 2\), which is positive. Therefore, it confirms the presence of a local minimum at \(z = 0\).
Quotient Rule
The Quotient Rule is a method for finding the derivative of a quotient or division of two functions, important for solving our given exercise. When dealing with a function like \(g(z) = \frac{z^2}{1+z^2}\), we apply this rule.
This rule states:
Inserting these into the Quotient Rule gives:
This rule states:
- If \(f(z) = \frac{u(z)}{v(z)}\) is a function where both \(u(z)\) and \(v(z)\) are differentiable, then the derivative \(f'(z)\) is given by:
- \[f'(z) = \frac{u'(z)v(z) - u(z)v'(z)}{[v(z)]^2}\]
Inserting these into the Quotient Rule gives:
- \[g'(z) = \frac{2z(1+z^2) - z^2(2z)}{(1+z^2)^2} = \frac{2z}{(1+z^2)^2}\]
