a) To find the x-coordinate of the relative maximum and minimum points, we need to find the critical points of the function. The critical points occur when the derivative of the function is equal to zero or undefined.
First, let's find the derivative of f(x):
f'(x) = 2(1/(x^2+3))(2x) - 1
To find the critical points, we set f'(x) equal to zero and solve for x:
2(1/(x^2+3))(2x) - 1 = 0
Simplifying the equation, we get:
4x/(x^2+3) - 1 = 0
Multiplying both sides by (x^2+3) to get rid of the fraction, we have:
4x - (x^2+3) = 0
Rearranging the equation, we get:
x^2 - 4x + 3 = 0
Factoring the quadratic equation, we have:
(x - 1)(x - 3) = 0
Setting each factor equal to zero and solving for x, we find that the critical points are x = 1 and x = 3.
Now, we need to test the intervals between the critical points and the endpoints of the domain to determine if they are relative maximum or minimum points.
When x < -3:
f'(x) = 2(1/(x^2+3))(2x) - 1 < 0
This means that f(x) is decreasing in this interval.
When -3 ≤ x < 1:
f'(x) = 2(1/(x^2+3))(2x) - 1 > 0
This means that f(x) is increasing in this interval.
When 1 < x < 3:
f'(x) = 2(1/(x^2+3))(2x) - 1 < 0
This means that f(x) is decreasing in this interval.
When 3 < x ≤ 5:
f'(x) = 2(1/(x^2+3))(2x) - 1 > 0
This means that f(x) is increasing in this interval.
Therefore, at x = 1 we have a relative minimum point, and at x = 3 we have a relative maximum point.
b) To find the inflection points of f, we need to find where the second derivative of f(x) changes sign or is equal to zero. Let's find the second derivative:
f''(x) = [4(2x)(x^2 + 3) - (2(2x))^2] / (x^2 + 3)^2
Simplifying, we have:
f''(x) = (8x^3 + 24x - 8x^2 - 12) / (x^2 + 3)^2
To find the inflection points, we set f''(x) equal to zero and solve for x:
8x^3 + 24x - 8x^2 - 12 = 0
Unfortunately, this equation cannot be easily solved analytically. However, we can use numerical methods or graphing software to approximate the inflection points.
c) To find the absolute maximum value of f(x), we need to consider the function values at the critical points and the endpoints of the given domain (-3 ≤ x ≤ 5).
We evaluate f(x) at the critical points and endpoints:
f(-3) = 2ln((-3)^2 + 3) - (-3) ≈ 4.7
f(1) = 2ln((1)^2 + 3) - (1) ≈ 0.61
f(3) = 2ln((3)^2 + 3) - (3) ≈ 2.48
f(5) = 2ln((5)^2 + 3) - (5) ≈ 0.37
Comparing the function values, we can see that the absolute maximum value is approximately 4.7, which occurs at x = -3.