# I needed help with these FRQ in my APCalc course. Any help or walkthrough would be extremely helpful - thanks in advance.

Let f be the function given by f(x)=3ln((x^2)+2)-2x with the domain [-2,4].
(a) Find the coordinate of each relative maximum point and each relative minimum point of f. Justify your answer.
(b) Find the x-coordinate of each point of inflection of the graph of f.
(c) Find the absolute maximum value of f(x).

Thanks again...

## the inner-most brackets are not needed, so

f(x)=3ln(x^2 + 2) - 2x

you should know how to differentiate a log function, so

f ' (x) = 3(2x)/(x^2 + 2) - 2

= 0 for max/min
6x/(x^2 + 2) = 2
3x = x^2 + 2
x^2 - 3x + 2 = 0
(x-2)(x-1) = 0
x = 2 or x = 1

when x = 2, f(2) = 3 ln(6) - 2
when x = 1 , f(1) = 3 ln(2) - 2

for pts of inflection, we need f '' (x) = 0
f ' (x) = 6x(x^2 + 2)^-1 - 2
f '' (x) = 6x(-1)(x^2 + 2)^-2 (2x) + 6(x^2 + 2)^-1
= 0

12x/(x^+2)^2 = 6/(x^2+2)
2x/(x^2 + 2) = 1
x^2 + 2 = 2x
x^2 - 2x + 2 = 0
which has no real solution, so there is no inflection point

for the absolutie max
evaluate f(-2), the left bounday
evaluate f(4) the right boundary
the above f(2) and the above f(1)
determine the largest of these values

## f'= 6x/(x^2+2) -2

set to zero
6x=2x^2+4
put in standard form
x^2-3x+2=0
(x-2)(x-1)=0
x=2, and x=1
now test each for second derivative.
f"=6/(x^2+2)-12x^2/(x^2+2)^2
at x=1
f"=6/3-12/9 which is positive, so at x=1 it is a rel min
at x=2
f"=6/4 -12 (4/36)=-3+1.5 which is negative, so it is a relative maximum.
to find the y value at each point, put in x in the basic function and compute

absolute maximum occurs at x=-2, (graph the funcion)

## Sure! I'll provide a step-by-step walkthrough for each part of the question:

(a) To find the relative maximum and minimum points of the function f(x), we need to find the critical points and determine their nature.

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

2. Set the derivative equal to zero to find the critical points:
(3 / (x^2 + 2)) * 2x - 2 = 0

3. Solve the equation for x:
(3 / (x^2 + 2)) * 2x = 2
3x / (x^2 + 2) = 1
3x = x^2 + 2
x^2 - 3x + 2 = 0

(x - 1)(x - 2) = 0

5. Solve for x:
x = 1 or x = 2

6. Now, we need to determine the nature of each critical point. To do this, we will analyze the second derivative of f(x).

7. Find the second derivative of f(x):
f''(x) = d/dx[ (3 / (x^2 + 2)) * 2x - 2]
f''(x) = -6x / (x^2 + 2)^2

8. Test the values of x = 1 and x = 2 in the second derivative expression.

For x = 1:
f''(1) = -6(1) / (1^2 + 2)^2 = -6/9

For x = 2:
f''(2) = -6(2) / (2^2 + 2)^2 = -12/16

9. Based on the sign of f''(x), we can determine the nature of the critical points:

At x = 1: Since f''(1) = -6/9 < 0, the point (1, f(1)) is a relative maximum.

At x = 2: Since f''(2) = -12/16 < 0, the point (2, f(2)) is also a relative maximum.

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

1. Find the second derivative of f(x), which we already calculated above:
f''(x) = -6x / (x^2 + 2)^2

2. Set the second derivative equal to zero to find the possible points of inflection:
-6x / (x^2 + 2)^2 = 0

3. Solve the equation for x:
-6x = 0
x = 0

The point of inflection exists at x = 0.

(c) To find the absolute maximum value of f(x) within the domain [-2, 4], we will evaluate the function f(x) at both the endpoints and critical points and determine the maximum value.

1. Evaluate f(x) at the critical points and endpoints:
f(-2), f(4), f(1), f(2), and f(0).

2. Compute the respective values:
f(-2) = 3ln((-2^2) + 2) - 2(-2)
f(-2) = 3ln(4) + 4

f(4) = 3ln((4^2) + 2) - 2(4)
f(4) = 3ln(18) - 8

f(1) = 3ln((1^2) + 2) - 2(1)
f(1) = 3ln(3) - 2

f(2) = 3ln((2^2) + 2) - 2(2)
f(2) = 3ln(6) - 4

f(0) = 3ln((0^2) + 2) - 2(0)
f(0) = 3ln(2)

3. Compare the values obtained and identify the absolute maximum value.

The maximum value of the function f(x) within the given domain is the largest value among f(-2), f(4), f(1), f(2), and f(0).

I hope this helps you understand how to solve these FRQs! Let me know if you have any further questions.