Consider the function below.
f(x) = 9 + 4x2 − x4
(a) Find the interval of increase. (Enter your answer using interval notation.)
Find the interval of decrease. (Enter your answer using interval notation.)
(b) Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
(c) Find the inflection points.
(x, y) =
(smaller x-value)
(x, y) =
(larger x-value)
Find the interval where the graph is concave upward. (Enter your answer using interval notation.)
Find the interval where the graph is concave downward. (Enter your answer using interval notation.)
Thank you so much, and if you could show me how so I can use the same method for my other problems, I would greatly appreciate it. Thanks!

You'll need the first and 2nd derivatives to do this problem.
f(x) = -x^4 + 4x^2 + 9
f'(x)=-4x^3+8x
f"(x)=-12x^2+8
The function is increasing where the 1st derivative is positive and decreasing where it is negative
It is concave down where the 2nd derivative is negative, concave up where it is positive
The critical points are where the 1st derivative = 0
-4x^3+8x=0
-4x(x^2-2)=0
either -4x=0 or x^2-2=0 which means x=0 or x=+/-√2
to see how the function behaves, look at values at the critical points, and also around the critical points
So let's look at...
.f(-2)=9...............(-2,9).......... f '(-2)=16............f "(-2) is negative
.f(-√2)=13...........(-√2,13).......f '(-√2)=0............f " is -
.f(-1)=12............ (-1,12).........f '(-1)=-4............. f " is -
.f(0)=9.................(0,9)............f '(0)=0................f " is +
f(1)=12..............(1,12)......... f '(1)=4................f " is -
f(√2)=13............(√2,13)........f '(√2)=0.............f " is -
.f(2)=9................(2,9) ...........f '(2)=-16.............f " is -
Back to the critical points.
When x<-√2, the 1st derivative is positive, meaning the function is increasing
When -√2<x<0 the 1st derivative is negative, meaning the function is decreasing
When 0<x<√2 the 1st derivative is positive, increasing
When x>√2 the 1st derivative is negative, decreasing
So at (-√2,13) it is increasing to the left and decreasing to the right - local max
At (0,9) decreasing to the left and increasing to the right - local min
At (√2,13) it is increasing to the left and decreasing to the right - local max
(NOTE: if you have a case where it is increasing or decreasing on both sides of a critical point, that is an inflection point - picture the graph of x^3 it always increases but at 0 it changes from curving down to curving up f(x)=x^3 , f'(x)=3x^2 it equals 0 at x=0 but is always positive the function always increases, f"(x)=6x which equals 0 at x=0 to the left it is negative and to the right it is positive)
It is mostly concave down except for a little bit around x=0 where it is concave up - it has two humps
The inflection points are where the 2nd derivative is 0 or
-12x^2+8=0
x^2=8/12=2/3
x=+/-√2/3=+/-0.82
2 increasing intervals and 2 decreasing intervals (so you'll need to write them with unions)
3 critical points
2 inflection points
I'll leave it to you to clean up since interval notation is a pain in these comment boxes
It might be a bit easier if you do a rough sketch with the points you find then indicate with an arrow for increasing/decreasing and curves for up or down