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)
To find the local maximum and minimum values of the function f(x) = x/(x^2+4), we can use both the first and second derivative tests.
First, let's find the first derivative of f(x) and set it equal to zero to find the critical points:
f'(x) = [(x^2+4)(1) - x(2x)] / (x^2+4)^2
Simplifying, we get:
f'(x) = (4 - 2x^2) / (x^2+4)^2
Next, let's set f'(x) = 0 and solve for x to find the critical points:
(4 - 2x^2) / (x^2+4)^2 = 0
4 - 2x^2 = 0
2x^2 = 4
x^2 = 2
Taking the square root of both sides, we get:
x = ±√2
So, the critical points are x = √2 and x = -√2.
Now, let's find the second derivative of f(x):
f''(x) = [(2x)(x^2+4)^2 - (4 - 2x^2)(2(x^2+4)(2x))] / (x^2+4)^4
Simplifying, we get:
f''(x) = (4x^3 + 24x) / (x^2+4)^3
Now, let's evaluate the second derivative at the critical points:
f''(√2) = (4(√2)^3 + 24(√2)) / ((√2)^2+4)^3
= (8√2 + 24√2) / (2+4)^3
= (32√2) / 216
= (4√2) / 27
f''(-√2) = (4(-√2)^3 + 24(-√2)) / ((-√2)^2+4)^3
= (-8√2 - 24√2) / (2+4)^3
= (-32√2) / 216
= (-4√2) / 27
Now, let's determine the nature of the critical points using both the first and second derivative tests:
1. First Derivative Test:
- If f'(x) changes sign from negative to positive at a critical point x, then that critical point corresponds to a local minimum.
- If f'(x) changes sign from positive to negative at a critical point x, then that critical point corresponds to a local maximum.
To perform the first derivative test, let's look at the intervals between the critical points:
Interval between -√2 and √2: Let's choose a test point, for example, x = 0. Plugging it into f'(x), we get:
f'(0) = (4 - 2(0)^2) / (0^2+4)^2
= 4/16
= 1/4
Since f'(0) is positive, f(x) is increasing in this interval. Therefore, there is a local minimum at x = -√2 and a local maximum at x = √2.
2. Second Derivative Test:
- If f''(x) > 0 at a critical point x, then that critical point corresponds to a local minimum.
- If f''(x) < 0 at a critical point x, then that critical point corresponds to a local maximum.
From the second derivative calculations:
f''(√2) = (4√2) / 27 > 0
f''(-√2) = (-4√2) / 27 < 0
Therefore, we have further confirmation that there is a local minimum at x = -√2 and a local maximum at x = √2.
In conclusion, according to both the first and second derivative tests, the function f(x) = x/(x^2+4) has a local minimum at x = -√2 and a local maximum at x = √2.
Regarding which method I prefer, it depends on the specific situation and the complexity of the function. The first derivative test is generally simpler to compute but may not provide sufficient information in some cases. The second derivative test provides more information about the nature of the critical points but requires the calculation of the second derivative. It is often useful to use both methods to confirm the results.