Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

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. $$ g(z)=\frac{z^{2}}{1+z^{2}} $$

Step-by-Step Solution

Answer

The critical point at \( z = 0 \) is a local minimum.

Step 1: Find the first derivative

To find the critical points, we need to find the first derivative of the function \( g(z) = \frac{z^2}{1+z^2} \). Use the quotient rule, \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). So, we let \( u = z^2 \) and \( v = 1 + z^2 \). We find \( u' = 2z \) and \( v' = 2z \). The first derivative is: \[ g'(z) = \frac{(2z)(1+z^2) - (z^2)(2z)}{(1+z^2)^2} = \frac{2z + 2z^3 - 2z^3}{(1+z^2)^2} = \frac{2z}{(1+z^2)^2} \].

Step 2: Set the first derivative to zero

Set \( g'(z) = \frac{2z}{(1+z^2)^2} = 0 \) to find the critical points. Since the denominator is always positive, the critical points occur when the numerator is zero. Thus, \( 2z = 0 \) which gives \( z = 0 \).

Step 3: Apply the First Derivative Test

Check the sign of \( g'(z) \) around the critical point \( z = 0 \). Choose test points: \(-1\) and \(1\). - For \( z = -1 \), \( g'(-1) = \frac{-2}{2^2} = -\frac{1}{2} \), indicating \( g'(z) < 0 \).- For \( z = 1 \), \( g'(1) = \frac{2}{2^2} = \frac{1}{2} \), indicating \( g'(z) > 0 \).Thus, the function decreases to \( z = 0 \) and increases after \( z = 0 \), indicating a local minimum at \( z = 0 \).

Step 4: Find the second derivative

Compute the second derivative \( g''(z) \). The first derivative is \( g'(z) = \frac{2z}{(1+z^2)^2} \). Use the quotient rule or simplify: let \( f(z) = z \) and \( h(z) = (1+z^2)^2 \). The quotient rule gives \[ g''(z) = \frac{(2)(1+z^2)^2 - (2z)(4z(1+z^2))}{(1+z^2)^4} \].Simplify to find critical concavity at \( z = 0 \).

Step 5: Evaluate the second derivative at the critical point

Evaluate \( g''(0) \):\[ g''(0) = \frac{2(1+0^2)^2 - (0)(4\cdot0(1+0^2))}{(1+0^2)^4} = \frac{2}{1} = 2 \].Since \( g''(0) > 0 \), the function is concave up at \( z = 0 \), confirming a local minimum.

Key Concepts

First Derivative Test

The First Derivative Test is a classic method in calculus used to determine whether a critical point is a local maximum, a local minimum, or neither. Critical points are where the first derivative of a function equals zero or is undefined.
Here's how it works:

• Find the first derivative of the function.
• Set the derivative to zero to find the critical points.
• Choose test points around each critical point to evaluate the sign of the derivative.

To apply this test, you need to determine whether the function increases or decreases on either side of these critical points. For the function \( g(z) = \frac{z^2}{1+z^2} \), we found that \( z = 0 \) is a critical point. Evaluating just to the left and right of this point:- For \( z = -1 \), \( g'(z) \) was negative, indicating a decreasing function.- For \( z = 1 \), \( g'(z) \) was positive, indicating an increasing function.

This means that the function transitions from decreasing to increasing at \( z = 0 \), suggesting a local minimum there.

Second Derivative Test

The Second Derivative Test provides another means to classify critical points by using concavity. It requires the calculation of the second derivative of the function.Here's the procedure:

• Calculate the second derivative of the function.
• Substitute the critical points into the second derivative.
• Evaluate the sign of the second derivative at these points:
• If \( g''(z) > 0 \): The function is concave up, indicating a local minimum.
• If \( g''(z) < 0 \): The function is concave down, indicating a local maximum.
• If \( g''(z) = 0 \): The test is inconclusive.

Applying this to our function \( g(z) = \frac{z^2}{1+z^2} \), after finding \( g''(z) \) and substituting \( z = 0 \), we get \( g''(0) = 2 \). Since \( g''(0) > 0 \), the function is concave up at this point, confirming it as a local minimum.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative measures how a function's output changes as its input changes. It's like asking, "How fast is something changing at a specific point?"
For the function \( g(z) = \frac{z^2}{1+z^2} \), we used differentiation to find \( g'(z) \), the first derivative of the function. This involves applying the differentiation rules, such as:

• The Power Rule: For \( f(x) = x^n \), the derivative is \( nx^{n-1} \).
• The Quotient Rule: For a function \( f(x) = \frac{u}{v} \), the derivative is \( \frac{u'v - uv'}{v^2} \).

Differentiation helps us find the slope of the tangent line to the curve at any point, which leads us to determine critical points where the slope is zero.
In short, differentiation is the tool that sets us on the path to using the First and Second Derivative Tests to classify critical points effectively.