Question

f(x) = x4 - 32x2 + 7
(a) Find the intervals on which f is increasing. (Enter the interval that contains smaller numbers first.)
( , ) ∪ ( , )

Find the intervals on which f is decreasing. (Enter the interval that contains smaller numbers first.)
( , ) ∪ ( , )

(b) Find the local minimum and maximum values of f.
(min)
(max)

(c) Find the inflection points.
( , ) (smaller x value)
( , ) (larger x value)

Find the intervals on which f is concave up. (Enter the interval that contains smaller numbers first.)
( , ) ∪ ( , )

Find the interval on which f is concave down.
( , ) ...?

Answer 1

f is increasing if f '(x) = 4x^3 - 64x >0, so it is the same of  x(x^2 -16)>0, implies x>0 or (x-4)(x+4)>0, so x> + or -4
the answer is (-4 , 0) ∪ (4 , infinity)

f is decreasing if f '(x) = 4x^3 - 64x <0, so it is the same of  x(x^2 -16)<0, implies x<0 or (x-4)(x+4)<0, so x< + or -4
the answer is (- infinity, -4) ∪ (0 , 4)

the local minimum and maximum
f '(x) =0, impolies x=+ or -4, or x=0
or f'(o)=0,  and  f'(- 4)=f'(4)= 0, 
M(-4, 0) or M(4, 0) or M(0,0)

inflection points can be found by solving f '' (x)=12x^2 - 64 =0
x=+ or - 4sqrt(3) / 3
so the inflection point is  and M(- 4sqrt(3) / 3, f'' ( -4sqrt(3))  (smaller x value), and M(4sqrt(3) / 3, f'' (4sqrt(3)) (larger x value)

f is concave up if f ''>0
it means 12x^2 - 64>0, so the interval is (- infinity, -4sqrt(3) U (4sqrt(3), infinity)

f is concave down if f ''<0
it means 12x^2 - 64<0 so the interval is (-4sqrt(3)) U (4sqrt(3))