Let f be the function given by

f(x)=2ln(x^2+3)-x with domain -3 is less than or equal to x which is less than or equal to 5
a) Find the x-coordinate of each relative maximum point and each relative minimum point of f. Justify your answer.
b) Find the x-coordinate of each inflection point of f.
c) Find the absolute maximum value of
f(x).

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.

a) To find the x-coordinate of each relative maximum and minimum point of f, we need to find the critical points of the function.

Step 1: Find the derivative of f(x).
f'(x) = (d/dx)(2ln(x^2+3)-x)
= 2(1/(x^2+3))(2x) - 1
= 4x/(x^2+3) - 1

Step 2: Set f'(x) equal to 0 to find the critical points.
4x/(x^2+3) - 1 = 0
4x = x^2 + 3
x^2 - 4x + 3 = 0

Step 3: Factor or use the quadratic formula to solve for x.
(x-1)(x-3) = 0
x = 1 or x = 3

Step 4: Check the second derivative to determine if the critical points are relative maximum or minimum points.
To find the second derivative f''(x), differentiate f'(x).
f''(x) = (d/dx)(4x/(x^2+3) - 1)
= [4(x^2+3) - 8x^2]/(x^2+3)^2
= (4-4x^2)/(x^2+3)^2

To determine the sign of f''(x) at x=1 and x=3, we substitute these values into f''(x).

f''(1) = (4-4(1)^2)/(1^2+3)^2 = 0
f''(3) = (4-4(3)^2)/(3^2+3)^2 = -8/72 = -1/9

Since f''(1) = 0 and f''(3) < 0, x=1 is not a critical point and x=3 is a relative maximum point.

Therefore, the x-coordinate of the relative maximum point is 3.

b) To find the x-coordinate of each inflection point, we need to find the x-values where the concavity of the function changes.

Step 1: Find the second derivative of f(x) obtained in part a.
f''(x) = (4-4x^2)/(x^2+3)^2

Step 2: Set f''(x) equal to 0 to find potential inflection points.
(4-4x^2)/(x^2+3)^2 = 0
4-4x^2 = 0
4x^2 = 4
x^2 = 1
x = 1 or x = -1

Step 3: Check the concavity of the function between these x-values.
Substitute the values x=1 and x=-1 into f''(x) to determine the sign.

f''(1) = (4-4(1)^2)/(1^2+3)^2 = 0
f''(-1) = (4-4(-1)^2)/((-1)^2+3)^2 = 0

Since f''(1) = 0 and f''(-1) = 0, there are no inflection points.

c) To find the absolute maximum value of f(x), we evaluate the function at the endpoints of the given domain (-3 and 5) as well as critical points found in part a.

f(-3) = 2ln((-3)^2+3) - (-3) = 2ln(12) + 3
f(5) = 2ln((5)^2+3) - (5) = 2ln(28) - 5
f(3) = 2ln((3)^2+3) - (3) = 2ln(12) - 3

To find the largest value among these three values, we can compare them.

f(-3) ≈ 4.365
f(5) ≈ 5.459
f(3) ≈ 0.847

The absolute maximum value of f(x) is approximately 5.459.

To solve these questions, we can follow these steps:

a) To find the relative maximum and minimum points, we need to find the critical points where the derivative of the function is equal to zero or does not exist. Here's how you can find them:

1. Find the derivative of f(x):
f'(x) = 2(1/(x^2+3))(2x) - 1

2. Set the derivative equal to zero and solve for x:
2(1/(x^2+3))(2x) - 1 = 0

Simplifying, we get:
(4x)/(x^2+3) - 1 = 0
4x - (x^2 + 3) = 0
x^2 - 4x + 3 = 0

Factoring the quadratic equation, we get:
(x - 1)(x - 3) = 0

So the critical points are x = 1 and x = 3.

3. Now we need to check the endpoints of the given domain, which are x = -3 and x = 5.

4. Finally, we evaluate the function at these critical points and endpoints to determine the relative maximum and minimum points.

Evaluate f(-3):
f(-3) = 2ln((-3)^2 + 3) - (-3)
= 2ln(12) + 3
≈ 4.48

Evaluate f(1):
f(1) = 2ln(1^2 + 3) - 1
= 2ln(4) - 1
≈ 1.77

Evaluate f(3):
f(3) = 2ln(3^2 + 3) - 3
= 2ln(12) - 3
≈ 2.48

Evaluate f(5):
f(5) = 2ln(5^2 + 3) - 5
= 2ln(28) - 5
≈ 1.08

Comparing these values, we find that the relative maximum point is (3, 2.48) and the relative minimum point is (1, 1.77).

b) To find the inflection points, we need to find the points where the second derivative changes sign. Here's how you can find them:

1. Find the second derivative of f(x):
f''(x) = (d/dx)[2(1/(x^2+3))(2x) - 1]

Simplifying, we get:
f''(x) = -4(x^2-1)/(x^2+3)^2

2. Set the second derivative equal to zero and solve for x:
-4(x^2-1)/(x^2+3)^2 = 0

The numerator is zero when x = -1 or x = 1, but these values are outside the given domain.

3. Now we need to check the values near x = -1 and x = 1 to see if there is a sign change in the second derivative.

Evaluate f''(-2):
f''(-2) = -4((-2)^2-1)/((-2)^2+3)^2
= -3/25

Evaluate f''(0):
f''(0) = -4((0)^2-1)/((0)^2+3)^2
= 1/9

Evaluate f''(2):
f''(2) = -4((2)^2-1)/((2)^2+3)^2
= -3/25

Since the sign of the second derivative does not change near x = -1 and x = 1, there are no inflection points in the given domain.

c) To find the absolute maximum value of f(x), we need to compare the function values at the critical points and endpoints of the given domain.

Evaluate f(-3), f(1), f(3), and f(5) from the previous step.

Comparing these values, we find that the absolute maximum value of f(x) is approximately 4.48, which occurs at x = -3.

f'(x) = 2(2x)/(x^2 + 1) - 1

= 0 for max/min

for this I got x = 2 ± √5
These lie within your domain so

You will have to find
f(-3), f(2+√5)), f(2-√5) and f(5) to see which is the maximum

for f''(x) I got (4(x^2 + 1) - 4x(2x))/(x^2 + 2)^2

setting this equal to zero, I got x = ± √2

sub back in the original to find the two inflection points.
f(-3)