Question

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...

Answer 1

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)

Answer 2

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