find the local maximum and minimum values of f using both the first and second derivative tests. which method do you prefer? f(x) = x/(x^2+4)

gjcc

To find the local maximum and minimum values of a function, f(x), we can use the first and second derivative tests. These tests involve finding the critical points of the function (where the derivative is zero or undefined) and then determining whether each critical point corresponds to a local maximum or minimum using the first and second derivative tests.

Let's start by finding the critical points of the function f(x) = x/(x^2 + 4).

1. Find the first derivative of f(x):
To do this, you can use the quotient rule. The quotient rule states that if you have a function of the form g(x)/h(x), where g(x) and h(x) are both differentiable functions, the derivative is given by (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2.

For f(x) = x/(x^2 + 4), let's denote g(x) = x and h(x) = (x^2 + 4). The derivative can be calculated as follows:

f'(x) = [(x^2 + 4) * 1 - x * 2x] / (x^2 + 4)^2
= (x^2 + 4 - 2x^2) / (x^2 + 4)^2
= (4 - x^2) / (x^2 + 4)^2

2. Set the first derivative equal to zero and solve for x:
To find the critical points, we set f'(x) = 0 and solve for x:

(4 - x^2) / (x^2 + 4)^2 = 0

Since the numerator, 4 - x^2, is already equal to zero, we get:

4 - x^2 = 0
x^2 = 4
x = ±2

So, we have two critical points: x = 2 and x = -2.

Now, let's move on to using the second derivative test to determine whether these critical points correspond to local maximum or minimum.

3. Find the second derivative of f(x):
To do this, we differentiate f'(x) obtained earlier:

f''(x) = -2x / (x^2 + 4)^2

4. Evaluate the second derivative at the critical points:
Substituting x = 2 into f''(x), we get:

f''(2) = -2(2) / (2^2 + 4)^2
= -4 / (4 + 4)^2
= -4 / (8)^2
= -4 / 64
= -1/16

Substituting x = -2 into f''(x), we get:

f''(-2) = -2(-2) / ((-2)^2 + 4)^2
= 4 / (4 + 4)^2
= 4 / (8)^2
= 4 / 64
= 1/16

5. Determine local maximum and minimum:
According to the second derivative test:
- If f''(x) > 0, then the function has a local minimum at x.
- If f''(x) < 0, then the function has a local maximum at x.
- If f''(x) = 0, the test is inconclusive.

Since f''(2) = -1/16 < 0, the critical point x = 2 corresponds to a local maximum.
Since f''(-2) = 1/16 > 0, the critical point x = -2 corresponds to a local minimum.

To answer your question about my preference, both methods can be used to find local maximum and minimum values. The first derivative test allows you to determine the intervals where the function is increasing or decreasing, and the second derivative test helps you determine the concavity of the function at critical points. Both methods provide valuable information about the behavior of the function. It is recommended to use both methods to confirm your findings and ensure accuracy in your analysis.